All Exams >
EmSAT Achieve >
Physics for EmSAT Achieve >
All Questions
All questions of Electric Charge for EmSAT Achieve Exam
Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be
- a)Zero
- b)Along OF
- c)Along OC
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be
a)
Zero
b)
Along OF
c)
Along OC
d)
None of these
|
Dr Manju Sen answered |
The charges will be balanced by their counterparts on the opposite side. So, eventually the charges remaining will be 2q and b and 3q on D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
- a)9.6 x 104 C
- b)1.9 x 105 C
- c)3.8 x 105 C
- d)4.8 x 104 C
Correct answer is option 'B'. Can you explain this answer?
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
a)
9.6 x 104 C
b)
1.9 x 105 C
c)
3.8 x 105 C
d)
4.8 x 104 C
|
Krishna Iyer answered |
- He atom has 2 electrons.
- So, 1 mole of He has 2*N(N is Avogadro's no.) electrons.
- Then total -ve charge in 1 mole He gas is = 2 * N * charge of 1 electron
= 2 * (6.022 * 1023) * (1.6 * 10-19) = 1.92 * 105 C
Can you explain the answer of this question below:Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
- A:
positive
- B:
Zero
- C:
positive or negative depending on which end is grounded
- D:
negative
The answer is b.
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
positive
Zero
positive or negative depending on which end is grounded
negative
|
Neha Sharma answered |
The charge on the rod is shared between the rod and the sphere when they are in contact with each other. However, on grounding the charge will flow to the earth and the charge on the sphere becomes zero
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R
- a)Rλ/ε0
- b)2Rλ/ε0
- c)λ/ε0
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R
a)
Rλ/ε0
b)
2Rλ/ε0
c)
λ/ε0
d)
none of the above
|
|
Nandini Patel answered |
The total charge on the body If there are n1 electrons and n2 protons will be |n1-n2|.
Because whichever of the two will be lower in number, will be neautralized by the other and the left electrons/ protons will be the reason for the charge on the body.
A negative point charge placed at the point A is 
- a) In stable equilibrium along x-axis
- b)In unstable equilibrium along y-axis
- c)in stable equilibrium along y-axis
- d)in unstable equilibrium along x-axis
Correct answer is option 'C,D'. Can you explain this answer?
A negative point charge placed at the point A is
a)
In stable equilibrium along x-axis
b)
In unstable equilibrium along y-axis
c)
in stable equilibrium along y-axis
d)
in unstable equilibrium along x-axis
|
EduRev Support answered |
If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d2U/dx2>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d2U/dx2<0.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude
a)
b)
c)
d)
|
Preeti Iyer answered |
Due to verifying electric field, it experiences an verifying force :-
F=QE=QE0sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax=QE0
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
mω2A=QE0
⇒A= QE0/mω2
F=QE=QE0sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax=QE0
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
mω2A=QE0
⇒A= QE0/mω2
Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero- a)10 cm. away from q and between the charge
- b)10 cm. away from 4q and out side the line joining them.
- c)20 cm. away from 4q and between the charge
- d)none
Correct answer is option 'C'. Can you explain this answer?
Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero
a)
10 cm. away from q and between the charge
b)
10 cm. away from 4q and out side the line joining them.
c)
20 cm. away from 4q and between the charge
d)
none
|
Hansa Sharma answered |
Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.
A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then- a)(Q/q) = (4/1)
- b)(Q/q) = (2/1)
- c)(Q/q) = (3/1)
- d)(Q/q) = (5/1)
Correct answer is option 'B'. Can you explain this answer?
A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then
a)
(Q/q) = (4/1)
b)
(Q/q) = (2/1)
c)
(Q/q) = (3/1)
d)
(Q/q) = (5/1)
|
Om Desai answered |
The force between q & Q-q is given by :-
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1
Can you explain the answer of this question below:If mica and woolen cloth are rubbed together, then mica gets- A:positively charged
- B:negatively charged
- C:remains neutral
- D:dual charged
The answer is a.
If mica and woolen cloth are rubbed together, then mica gets
A:
positively charged
B:
negatively charged
C:
remains neutral
D:
dual charged
|
Mira Sharma answered |
When Mika (quartz) and wooden cloth are rubbed together then Mika gets Positively charged. It's because of friction. Due to friction one gets Positively charged i.e Mika and other gets Negatively charged i.e Wooden cloth.
Which one of the following statement regarding electrostatics is wrong ?- a)Charge is quantized
- b)Charge is conserved
- c)There is an electric field near an isolated charge at rest
- d)A stationary charge produces both electric and magnetic fields
Correct answer is 'D'. Can you explain this answer?
Which one of the following statement regarding electrostatics is wrong ?
a)
Charge is quantized
b)
Charge is conserved
c)
There is an electric field near an isolated charge at rest
d)
A stationary charge produces both electric and magnetic fields
|
Nandita Ahuja answered |
Explanation:
Electrostatics is the study of electric charges at rest. It deals with the electric forces between charges, the electric field and potential, and the distribution of charges on conductors.
a) Charge is quantized:
The charge on a body or a particle is always a multiple of the elementary charge (1.6 × 10^-19 C). This means that charge is quantized, and we cannot have a fraction of an elementary charge. This is known as the law of conservation of charge.
b) Charge is conserved:
The total charge in a closed system is always conserved. This means that the net charge of a system cannot be created or destroyed; it can only be transferred from one object to another.
c) There is an electric field near an isolated charge at rest:
An isolated charged object creates an electric field around it. This electric field is a vector field that exerts a force on other charged objects in the vicinity of the charged object. The electric field is proportional to the charge and inversely proportional to the distance from the charged object.
d) A stationary charge produces both electric and magnetic fields:
This statement is incorrect. A stationary charge produces only an electric field, not a magnetic field. However, a moving charge produces both electric and magnetic fields.
Conclusion:
Hence, the correct option is (d), which is wrong.
Electrostatics is the study of electric charges at rest. It deals with the electric forces between charges, the electric field and potential, and the distribution of charges on conductors.
a) Charge is quantized:
The charge on a body or a particle is always a multiple of the elementary charge (1.6 × 10^-19 C). This means that charge is quantized, and we cannot have a fraction of an elementary charge. This is known as the law of conservation of charge.
b) Charge is conserved:
The total charge in a closed system is always conserved. This means that the net charge of a system cannot be created or destroyed; it can only be transferred from one object to another.
c) There is an electric field near an isolated charge at rest:
An isolated charged object creates an electric field around it. This electric field is a vector field that exerts a force on other charged objects in the vicinity of the charged object. The electric field is proportional to the charge and inversely proportional to the distance from the charged object.
d) A stationary charge produces both electric and magnetic fields:
This statement is incorrect. A stationary charge produces only an electric field, not a magnetic field. However, a moving charge produces both electric and magnetic fields.
Conclusion:
Hence, the correct option is (d), which is wrong.
In SI units, a unit of charge is called a- a)ampere
- b)milli coulomb
- c)milli ampere
- d)Coulomb
Correct answer is option 'D'. Can you explain this answer?
In SI units, a unit of charge is called a
a)
ampere
b)
milli coulomb
c)
milli ampere
d)
Coulomb
|
Ayush Joshi answered |
The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). It is a dimensionless quantity, sharing this aspect with the mole. A quantity of 1 C is equal to approximately 6.24 x 1018, or 6.24 quintillion.
In terms of SI base units, the coulomb is the equivalent of one ampere-second. Conversely, an electric current of A represents 1 C of unit electric charge carriers flowing past a specific point in 1 s. The unit electric charge is the amount of charge contained in a single electron.
In space of horizontal EF(E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position. 
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In space of horizontal EF(E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position.
a)
b)
c)
d)
|
|
Priyanka Sharma answered |
According to the work energy theorem we have
We+Wg=(1/2)mv2
We have work done by electrostatic force as
We=qElsinθ
and work done by the gravitational force as
Wg=mg(l−lcosθ)
Thus we get
qElsinθ+mg(l−lcosθ)=(1/2)mv2
Thus we get
mgsinθ+mgl−mglcosθ=(1/2)mv2
as θ=45o, we get
mgl=(1/2)mv2
also as v=ωl
we get
ω=√2g/l
We+Wg=(1/2)mv2
We have work done by electrostatic force as
We=qElsinθ
and work done by the gravitational force as
Wg=mg(l−lcosθ)
Thus we get
qElsinθ+mg(l−lcosθ)=(1/2)mv2
Thus we get
mgsinθ+mgl−mglcosθ=(1/2)mv2
as θ=45o, we get
mgl=(1/2)mv2
also as v=ωl
we get
ω=√2g/l
Can you explain the answer of this question below:A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to
- A:
x-1
- B:
x-3/2
- C:
x-2
- D:
x5/4
The answer is C.
A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to
x-1
x-3/2
x-2
x5/4
|
Lavanya Menon answered |
Electric field due to a charged ring in given by: -
at point p
∣E∣= KQx /(R2+x2)3/2 Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣E∣= K&x/(x2)3/2=KQ/x2=Eαx−2
so at a large distance, the ring behaves as a point particle.
at point p
∣E∣= KQx /(R2+x2)3/2 Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣E∣= K&x/(x2)3/2=KQ/x2=Eαx−2
so at a large distance, the ring behaves as a point particle.
Quarks are entities that have- a)Fractional charges, combination of these are present in protons and neutrons
- b)integer charges, combination of these are present in protons, electrons and neutrons
- c)fractional charges, combination of these are present in protons and electrons
- d)no charge, combination of these are present in protons and electrons
Correct answer is 'A'. Can you explain this answer?
Quarks are entities that have
a)
Fractional charges, combination of these are present in protons and neutrons
b)
integer charges, combination of these are present in protons, electrons and neutrons
c)
fractional charges, combination of these are present in protons and electrons
d)
no charge, combination of these are present in protons and electrons
|
Prabhakar Kumar Tiwari answered |
(a)quarks combine to form hadrons most of which are protons and neutrons.quarks are the only known particles whose charges are not integral multiples of their charges u can refer Google for detailed information
C
C
C
A charged particle of charge q and mass m is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time 't' seconds is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A charged particle of charge q and mass m is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time 't' seconds is
a)
b)
c)
d)
|
Vivek Rana answered |
Force on particle=F=qE
Hence, acceleration of particle=a=F/m=qE/m
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m
Kinetic energy=K=(1/2)mv2=(1/2)m(qEt/m)2
⟹K=E2q2t2/2m
Hence, acceleration of particle=a=F/m=qE/m
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m
Kinetic energy=K=(1/2)mv2=(1/2)m(qEt/m)2
⟹K=E2q2t2/2m
When a negatively charged conductor is connected to earth,- a)No charge flow occurs.
- b)Protons flow from the conductor to the earth
- c)Electrons flow from the earth to the conductor
- d)Electrons flow from the conductor to the earth
Correct answer is option 'D'. Can you explain this answer?
When a negatively charged conductor is connected to earth,
a)
No charge flow occurs.
b)
Protons flow from the conductor to the earth
c)
Electrons flow from the earth to the conductor
d)
Electrons flow from the conductor to the earth
|
|
Om Desai answered |
After earthing a positively charged conductor electrons flow from earth to conductor and if a negatively charged conductor is earthed then electrons flows from conductor to earth.
Four charges are placed each at a distance a from origin. The dipole moment of configuration is
- a)
- b)
- c)
- d)none
Correct answer is option 'A'. Can you explain this answer?
Four charges are placed each at a distance a from origin. The dipole moment of configuration is
a)
b)
c)
d)
none
|
|
Dr Manju Sen answered |
This question can be solved easily , if you have some knowledge about resolution of vectors.
all four charges are placed , each at a distance 'a ' from the origin.
it means distance between two charges = √{a² + a² } = √2a
We also know, dipole moment is the system of two equal magnitude but opposite nature charges . so, we have to divide the charges as shown in the figure for making dipoles .
Hence, there are three dipoles formed.
Let P = q(√2a)
then, P₁ = P , P₂ = P and P₃ = 2P {as shown in figure.}
now resolve the vectors P₁ ,P₂ and P₃ .
as shown in figure,
vertical components of P₁ and P₂ is cancelled .
and horizontal components of P₁ and P₂ is cancelled by horizontal component of P₃ .
and rest part of dipole = vertical component of P₃ = 2Psin45° j ['j' shows direction of net dipole moment]
hence, net dipole moment = 2q(√2a) × 1/√2 j
= 2qa j
all four charges are placed , each at a distance 'a ' from the origin.
it means distance between two charges = √{a² + a² } = √2a
We also know, dipole moment is the system of two equal magnitude but opposite nature charges . so, we have to divide the charges as shown in the figure for making dipoles .
Hence, there are three dipoles formed.
Let P = q(√2a)
then, P₁ = P , P₂ = P and P₃ = 2P {as shown in figure.}
now resolve the vectors P₁ ,P₂ and P₃ .
as shown in figure,
vertical components of P₁ and P₂ is cancelled .
and horizontal components of P₁ and P₂ is cancelled by horizontal component of P₃ .
and rest part of dipole = vertical component of P₃ = 2Psin45° j ['j' shows direction of net dipole moment]
hence, net dipole moment = 2q(√2a) × 1/√2 j
= 2qa j
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.- a)positive
- b)Zero
- c)positive or negative depending on which end is grounded
- d)negative
Correct answer is 'B'. Can you explain this answer?
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
a)
positive
b)
Zero
c)
positive or negative depending on which end is grounded
d)
negative
|
Amrutha Pillai answered |
The charge on the rod is shared between the rod and the sphere when they are in contact with each other. However, on grounding the charge will flow to the earth and the charge on the sphere becomes zero
Earthing or grounding means- a)placing the apparatus with an insulating stand, on the ground
- b)connecting to a green colored wire
- c)sharing the charges with the earth
- d)using green color to paint the body
Correct answer is option 'C'. Can you explain this answer?
Earthing or grounding means
a)
placing the apparatus with an insulating stand, on the ground
b)
connecting to a green colored wire
c)
sharing the charges with the earth
d)
using green color to paint the body
|
|
Preeti Iyer answered |
Earthing and Grounding are actually different terms for expressing the same concept. Ground or earth in a mains electrical wiring system is a conductor that provides a low impedance path to the earth to prevent hazardous voltages from appearing on equipment. In electrical engineering, ground or earth is the reference point in an electrical circuit from which voltages are measured, a common return path for electric current, or a direct physical connection to the earth.
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.- a)positive
- b)Zero
- c)positive or negative depending on which end is grounded
- d)negative
Correct answer is option 'B'. Can you explain this answer?
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
a)
positive
b)
Zero
c)
positive or negative depending on which end is grounded
d)
negative
|
|
Preeti Iyer answered |
The charge on the rod is shared between the rod and the sphere when they are in contact with each other. However, on grounding the charge will flow to the earth and the charge on the sphere becomes zero
Which of the following methods can be used to charge a metal sphere positively without touching it? Choose the best possible option:- a)Connect the positive terminal a battery and float the other end of the battery
- b)Rub it with a piece of fur
- c)Bring a negatively charged rod near the sphere and touch it to ground for a short while
- d)Rub it with a piece of silk
Correct answer is option 'C'. Can you explain this answer?
Which of the following methods can be used to charge a metal sphere positively without touching it? Choose the best possible option:
a)
Connect the positive terminal a battery and float the other end of the battery
b)
Rub it with a piece of fur
c)
Bring a negatively charged rod near the sphere and touch it to ground for a short while
d)
Rub it with a piece of silk
|
Ritu Singh answered |
This can be possible by induction, if a negative charge rod is placed near the sphere (at a distance not in touch) then it attracts positive charge inside the sphere towards its side and the negative charges collected on the other side and when the surface of the sphere connected to the earth than the negative charge goes to earth and the sphere remain positively charged.
Since rubbing the metal ball with fur and a piece of silk cloth involves touching, they can not be the correct answer.
Therefore, the correct answer is C.
Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charge starts on the x-axis at a large distance from O, moves along the +x-axis, passes through O and moves far away from O. Its acceleration a is taken as positive along its direction of motion. The particle's acceleration a is plotted against its x-coordinate. Which of the following best represents the plot ?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle with a negative charge starts on the x-axis at a large distance from O, moves along the +x-axis, passes through O and moves far away from O. Its acceleration a is taken as positive along its direction of motion. The particle's acceleration a is plotted against its x-coordinate. Which of the following best represents the plot ?
a)
b)
c)
d)
|
New Words answered |
Consider the attached free body diagram for the given scenario.
From coulomb's law,
∣F1∣=∣F2∣= qQ /4πεor2
In y-direction, forces cancel, Fy=0
In x-direction, forces add up. Also, it is directed in +ve x direction if the position of −q is along −ve x axis and in −ve x direction if the position of −q is along +ve x axis.
Hence,
Fx=2qQcosθ i /4πεor2 ∀ x<0
Fx=2qQcosθ i /4πεor2 ∀ x≥0
But cosθ =x/r
r= √x2+d2
Substituting in force equation, we get
Fx=2qQx i /4πεo( x2+d2)3/2 ∀ x<0
Fx=2qQx i /4πεo( x2+d2)3/2 ∀ x≥0
Now, a=F/m by Newton's second law. It is positive when x is negative and negative when x is positive. This is satisfied only in option (B).
Can you explain the answer of this question below:When we rub a glass rod with silk, thena)Negative charge is produced on silk but not charge on the glass rodb)Equal and similar charges are produced on the bothc)Positive charge is produced on the glass rod but no charge on the silkd)Equal but opposite charge are produced on the bothThe answer is d.
a)Negative charge is produced on silk but not charge on the glass rod
b)Equal and similar charges are produced on the both
c)Positive charge is produced on the glass rod but no charge on the silk
d)Equal but opposite charge are produced on the both
The answer is d.
|
|
Wahid Khan answered |
When glass rod is rubbed with silk electrons are transferred from glass to silk, therefore the mass of glass rod decreases
A silk cloth rubbed with a glass rod acquire a charge (-1.6 × 10 -19) C. Then the charge on the glass rod is- a)– 3.2 × 10 -19 C
- b)– 1.6 × 10 -19 C
- c)+1.6 × 10 -19 C
- d)3.2 × 10 -19 C
Correct answer is option 'C'. Can you explain this answer?
A silk cloth rubbed with a glass rod acquire a charge (-1.6 × 10 -19) C. Then the charge on the glass rod is
a)
– 3.2 × 10 -19 C
b)
– 1.6 × 10 -19 C
c)
+1.6 × 10 -19 C
d)
3.2 × 10 -19 C
|
Shaurya Kashaudhan answered |
The glass rod lose electon their for it aquire positive charge equal to magnitude of nigative charge
- a)so that the proton charge will be positive
- b)because like charges repel
- c)to conform to the conventions adopted for G and m in Newton’s law of gravitation
- d)by arbitrary convention
Correct answer is 'D'. Can you explain this answer?
|
Amrita Basak answered |
When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
a)
b)
c)
d)
|
|
Isha Rane answered |
He atom has 2 electrons. so 1 mole of He has 2*N(N is Avohadro's no.) electrons. then total -ve charge in 1 mole He gas is 2*N*charge of 1 electron. calculating we will get about 1.9*10^5 C.
Two identical point charges are placed at a separation of l.P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve ?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Two identical point charges are placed at a separation of l.P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve ?
a)
b)
c)
d)
|
Imk Pathsala answered |
At point P,
E=E1−E2 rightward
⟹E=(kq/x2)−(kq/(l−x)2) where k=1/4πϵo
⟹E=kq (l(l−2x)/x2(l−x)2)
From graph we can see that it can't be straight line and E=0 at x=l/2
At x=0 ,,E→∞ and at x=l, E→−∞
Hence, from the shown graphs the correct answer is (D).
If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be- a)100 N/C
- b)50 N/C
- c)200 N/C
- d)10 N/C
Correct answer is option 'B'. Can you explain this answer?
If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be
a)
100 N/C
b)
50 N/C
c)
200 N/C
d)
10 N/C
|
|
Preeti Iyer answered |
Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, E=F/q=100N/2C=50 N/C.
Hence, the value of field intensity will be 50 N/C.
In this case, F = 100 N and q = 2 C.
So, E=F/q=100N/2C=50 N/C.
Hence, the value of field intensity will be 50 N/C.
If body is positively or negatively charged leaves of electroscope will- a)diverge
- b)converge
- c)stay still
- d)shrink
Correct answer is option 'A'. Can you explain this answer?
If body is positively or negatively charged leaves of electroscope will
a)
diverge
b)
converge
c)
stay still
d)
shrink
|
|
Raghavendra Rane answered |
Explanation:
When a body is charged, it gains either positive or negative charge. This charge is transferred to the leaves of an electroscope. The leaves of an electroscope are made of a thin metal foil that is suspended from a metal rod. When the charged body is brought near the electroscope, the charge on the body induces a charge of the opposite sign in the leaves of the electroscope. This causes the leaves to either diverge or converge. The direction of divergence or convergence depends on the type of charge on the charged body.
When the body is positively charged, the leaves of the electroscope diverge. This is because the positive charge on the body repels the positive charge in the leaves, causing them to move away from each other.
When the body is negatively charged, the leaves of the electroscope also diverge. This is because the negative charge on the body repels the negative charge in the leaves, causing them to move away from each other.
In both cases, the movement of the leaves indicates the presence of a charge on the body. The greater the charge on the body, the greater the divergence of the leaves.
Conclusion:
Therefore, the correct answer is option 'A', which states that the leaves of the electroscope will diverge when the body is positively or negatively charged.
When a body is charged, it gains either positive or negative charge. This charge is transferred to the leaves of an electroscope. The leaves of an electroscope are made of a thin metal foil that is suspended from a metal rod. When the charged body is brought near the electroscope, the charge on the body induces a charge of the opposite sign in the leaves of the electroscope. This causes the leaves to either diverge or converge. The direction of divergence or convergence depends on the type of charge on the charged body.
When the body is positively charged, the leaves of the electroscope diverge. This is because the positive charge on the body repels the positive charge in the leaves, causing them to move away from each other.
When the body is negatively charged, the leaves of the electroscope also diverge. This is because the negative charge on the body repels the negative charge in the leaves, causing them to move away from each other.
In both cases, the movement of the leaves indicates the presence of a charge on the body. The greater the charge on the body, the greater the divergence of the leaves.
Conclusion:
Therefore, the correct answer is option 'A', which states that the leaves of the electroscope will diverge when the body is positively or negatively charged.
Electric potential is a -- a)Vector quantity
- b)Scalar quantity
- c) Neither vector Nor scalar
- d)Fictious quantity
Correct answer is option 'B'. Can you explain this answer?
Electric potential is a -
a)
Vector quantity
b)
Scalar quantity
c)
Neither vector Nor scalar
d)
Fictious quantity
|
|
Om Desai answered |
Electric potential is a pure Scalar quantity, The reason is as follows.
The Electric Potential is defined as the amount of work-done per unit positive charge to bring from infinity to that point under the influence of the primary charge only.
U=W/q
And work done is defined as the dot product of force and displacement which is a scalar quantity.
W=F.S
Thus Electric potential is a scalar quantity.
The Electric Potential is defined as the amount of work-done per unit positive charge to bring from infinity to that point under the influence of the primary charge only.
U=W/q
And work done is defined as the dot product of force and displacement which is a scalar quantity.
W=F.S
Thus Electric potential is a scalar quantity.
Both the leaves of the electroscope carry- a)Same charges
- b)zero charge
- c)opposite charge
- d)always negative charge
Correct answer is option 'A'. Can you explain this answer?
Both the leaves of the electroscope carry
a)
Same charges
b)
zero charge
c)
opposite charge
d)
always negative charge
|
|
Shreya Gupta answered |
When the metal terminal is touched with a charged object, the gold leaves spread apart in a 'V'. This is because some of the charge on the object is conducted through the terminal and metal rod to the leaves. Since they receive the same sign charge they repel each other and thus diverge.
The quantization of charge has little practical consequence and can be ignored at the- a)microscopic level
- b)macroscopic level
- c)all levels
- d)information given is insufficient to decide
Correct answer is 'B'. Can you explain this answer?
The quantization of charge has little practical consequence and can be ignored at the
a)
microscopic level
b)
macroscopic level
c)
all levels
d)
information given is insufficient to decide
|
|
Roshni Desai answered |
Gain or loss of charge in terms of few electrons each carrying very small charge is unobservable at macroscopic level.
A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -- a)For all values of H
- b)Only if H >
- c)Only if H <
- d)Only if H =
Correct answer is option 'B'. Can you explain this answer?
A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -
a)
For all values of H
b)
Only if H >
c)
Only if H <
d)
Only if H =
|
Sushil Kumar answered |
electric field due to non-conducting ring along its axis is given by,
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m. 
- a)The point P where the resultant field due to both is zero is on AB outside AB.
- b)The point P where the resultant field due to both is zero is on AB inside AB.
- c) If a positive charge is placed at P and displaced slightly along AB it will execute oscillations.
- d) If a negative charge is placed at P and displaced slightly along AB it will execute oscillation.
Correct answer is option 'A,D'. Can you explain this answer?
Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m.
a)
The point P where the resultant field due to both is zero is on AB outside AB.
b)
The point P where the resultant field due to both is zero is on AB inside AB.
c)
If a positive charge is placed at P and displaced slightly along AB it will execute oscillations.
d)
If a negative charge is placed at P and displaced slightly along AB it will execute oscillation.
|
Gopal Gupta answered |
A and C
The direction (q) of at point P due to uniformly charged finite rod will be -
- a)At angle 30° from x-axis
- b)45° from x-axis
- c)60° from x-axis
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The direction (q) of at point P due to uniformly charged finite rod will be -
a)
At angle 30° from x-axis
b)
45° from x-axis
c)
60° from x-axis
d)
None of these
|
Ciel Knowledge answered |
The angle suspended by the finite line charge at P=60°
So, the resultant electric field due to line charge will be at 60/2⇒30° since we can assume the charge concentrated at the centre of finite line charge.
So, the resultant electric field due to line charge will be at 60/2⇒30° since we can assume the charge concentrated at the centre of finite line charge.
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:- a)
- b)
- c)
- d)
Correct answer is 'A'. Can you explain this answer?
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
a)
b)
c)
d)
|
Tejas Desai answered |
He atom has 2 electrons. so 1 mole of He has 2*N(N is Avohadro's no.) electrons. then total -ve charge in 1 mole He gas is 2*N*charge of 1 electron. calculating we will get about 1.9*10^5 C.
If a body contains n1 electrons and n2 protons, the total amount of charge on the body is- a)(n2 x n1) e
- b)(n2 / n1) e
- c)(n2-n1)e
- d)(n2 + n1) e
Correct answer is option 'C'. Can you explain this answer?
If a body contains n1 electrons and n2 protons, the total amount of charge on the body is
a)
(n2 x n1) e
b)
(n2 / n1) e
c)
(n2-n1)e
d)
(n2 + n1) e
|
|
Hansa Sharma answered |
Correct Answer :- A
Explanation :
Line charge density = λ
radius, r=R
length of wire inside the sphere is 2 R
Now, λdl=dθ
for total charge enclosed by radius R, ∫(O to R)Q = ∫(O to R)λdl
Qinc = λ2R
Now, ϕ = Qinc/ε0
= λ2R/ε0
Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be- a)Zero
- b)4E
- c)E
- d)1/2E
Correct answer is option 'A'. Can you explain this answer?
Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be
a)
Zero
b)
4E
c)
E
d)
1/2E
|
|
Priyanka Sharma answered |
As the components of electric field intensity at diagonal are equal in magnitude and opposite in direction, thus the intensity of electric field will be zero.
Hence the correct answer would be option A.
Hence the correct answer would be option A.
A particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present. The path of particle- a)May be a straight line
- b)May be a circle
- c)May be a parabola
- d)May be a hyperbola
Correct answer is option 'A,C'. Can you explain this answer?
A particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present. The path of particle
a)
May be a straight line
b)
May be a circle
c)
May be a parabola
d)
May be a hyperbola
|
Rajesh Chatterjee answered |
Explanation:
When a particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present, its path can be determined by considering the forces acting on it.
Forces acting on the particle:
Possible paths of the particle:
Therefore, the correct answer is options A and C - the path of the particle may be a straight line or a parabola.
When a particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field are present, its path can be determined by considering the forces acting on it.
Forces acting on the particle:
- Gravitational force (Fg) = mg, where g is the acceleration due to gravity
- Electric force (Fe) = qE, where E is the electric field strength
Possible paths of the particle:
- Straight line: If the electric and gravitational forces are in the same direction, the particle will move in a straight line with constant acceleration. The path will be a straight line.
- Parabola: If the electric and gravitational forces are perpendicular to each other, the particle will move in a parabolic path. This happens when the particle is thrown at an angle to the electric field. The path will be a parabola.
- Circle: If the electric force is perpendicular to the gravitational force, the particle will move in a circular path. This happens when the particle is thrown in a direction perpendicular to the electric field. The path will be a circle.
Therefore, the correct answer is options A and C - the path of the particle may be a straight line or a parabola.
Magnitude of electric charge on a single electron is:- a)1.6 x 10-18 coulomb
- b)1.6 x 10-10 coulomb
- c)1.6 x 10-16 coulomb
- d)1.6 x 10-19 coulomb
Correct answer is option 'D'. Can you explain this answer?
Magnitude of electric charge on a single electron is:
a)
1.6 x 10-18 coulomb
b)
1.6 x 10-10 coulomb
c)
1.6 x 10-16 coulomb
d)
1.6 x 10-19 coulomb
|
|
Pooja Mehta answered |
One Coulomb is equal to 6.25 x 10^18 proton's worth or electron's worth of charge (depending on whether it's positive or negative charge). So one electron has -1.6 x 10^-19 Coulombs of charge and one proton has +1.6 x 10^-19 Coulombs of charge. We will use “C” to represent Coulombs.
Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length 'L' metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of weight lessness) Then the angle (q) between the two strings is- a)0º
- b)90º
- c)180º
- d)0º < q < 180º
Correct answer is option 'C'. Can you explain this answer?
Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length 'L' metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity (state of weight lessness) Then the angle (q) between the two strings is
a)
0º
b)
90º
c)
180º
d)
0º < q < 180º
|
Geetika Shah answered |
There is a condition of weightlessness in a satellite.
Therefore, mg=0
Due to electrostatic force of repulsion between the balls, the string would become horizontal.
Therefore, angle between the string =180
Therefore, mg=0
Due to electrostatic force of repulsion between the balls, the string would become horizontal.
Therefore, angle between the string =180
If mica and woolen cloth are rubbed together, then mica gets- a)positively charged
- b)negatively charged
- c)remains neutral
- d)dual charged
Correct answer is 'A'. Can you explain this answer?
If mica and woolen cloth are rubbed together, then mica gets
a)
positively charged
b)
negatively charged
c)
remains neutral
d)
dual charged
|
|
Nisha Pillai answered |
Woolen cloth has +ve charge and when mica is rubbed on it , mica gets +vely charged due to (conduction) - flow of charge from wool to mica
Quantization of charge means that- a)the charge on a body is a fraction of electronic charge, e
- b)The charge on a body is an integral multiple of electronic charge, e
- c)the net charge on a body is always zero
- d)the total charge on a body is always conserved
Correct answer is option 'B'. Can you explain this answer?
Quantization of charge means that
a)
the charge on a body is a fraction of electronic charge, e
b)
The charge on a body is an integral multiple of electronic charge, e
c)
the net charge on a body is always zero
d)
the total charge on a body is always conserved
|
|
Nandini Patel answered |
Charge quantization, then, means that charge cannot take any arbitrary values, but only values that are integral multiples of the fundamental charge (charge of proton/electron). For example, in a hydrogen ion, we usually denote it with a positive sign to indicate that there's one proton more than there are electrons.
. What is the electric field at the point of position vector 
The charge on a glass rod that has been rubbed with silk is called positive:- a)so that the proton charge will be positive
- b)because like charges repel
- c)to conform to the conventions adopted for G and m in Newton’s law of gravitation
- d)by arbitrary convention
Correct answer is 'D'. Can you explain this answer?
The charge on a glass rod that has been rubbed with silk is called positive:
a)
so that the proton charge will be positive
b)
because like charges repel
c)
to conform to the conventions adopted for G and m in Newton’s law of gravitation
d)
by arbitrary convention
|
Aravind Mehra answered |
When the point is on the diameter and away from the centre of hemisphere which is charged uniformly and positively, the component of electric field intensity parallel to the diameter cancel out. So the electric field is perpendicular to the diameter.
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
- a)8 mC
- b)2 mC
- c)5 mC
- d)7 μC
Correct answer is option 'B'. Can you explain this answer?
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
a)
8 mC
b)
2 mC
c)
5 mC
d)
7 μC
|
|
Srestha Chopra answered |
° with an electric field of magnitude 500 N/C. The dipole moment is 2 C.m. Find the torque experienced by the dipole.
The torque experienced by an electric dipole in an electric field is given by the formula:
τ = p E sinθ
where τ is the torque, p is the dipole moment, E is the electric field, and θ is the angle between the dipole moment and the electric field.
Substituting the given values, we get:
τ = (2 C.m) × (500 N/C) × sin30°
τ = (2 C.m) × (500 N/C) × 0.5
τ = 500 N.m
Therefore, the torque experienced by the dipole is 500 N.m.
The torque experienced by an electric dipole in an electric field is given by the formula:
τ = p E sinθ
where τ is the torque, p is the dipole moment, E is the electric field, and θ is the angle between the dipole moment and the electric field.
Substituting the given values, we get:
τ = (2 C.m) × (500 N/C) × sin30°
τ = (2 C.m) × (500 N/C) × 0.5
τ = 500 N.m
Therefore, the torque experienced by the dipole is 500 N.m.
Chapter doubts & questions for Electric Charge - Physics for EmSAT Achieve 2025 is part of EmSAT Achieve exam preparation. The chapters have been prepared according to the EmSAT Achieve exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for EmSAT Achieve 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Electric Charge - Physics for EmSAT Achieve in English & Hindi are available as part of EmSAT Achieve exam.
Download more important topics, notes, lectures and mock test series for EmSAT Achieve Exam by signing up for free.
Physics for EmSAT Achieve
215 videos|402 docs|221 tests
|
