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All questions of Electrostatics for Airforce X Y / Indian Navy SSR 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.
Three capacitors, each of capacitance C = 3 mF, are connected as shown in the figure. The equivalent capacitance between points P and S is
- a)9 μF
- b)6 μF
- c)3 μF
- d)1 μF
Correct answer is option 'A'. Can you explain this answer?
Three capacitors, each of capacitance C = 3 mF, are connected as shown in the figure. The equivalent capacitance between points P and S is
a)
9 μF
b)
6 μF
c)
3 μF
d)
1 μF
|
|
Dr Manju Sen answered |
If P is at positive potential, then Q is at negative potential and R is at positive potential. The system therefore reduces to 3 capacitors in parallel. C= 9μF
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
Electric flux is a _______ quantity- a)vector
- b)always negative
- c)Scalar
- d)always positive
Correct answer is option 'C'. Can you explain this answer?
Electric flux is a _______ quantity
a)
vector
b)
always negative
c)
Scalar
d)
always positive
|
Rajeev Saxena answered |
For transport phenomena, flux is a vector quantity, describing the magnitude and direction of the flow of a substance or property. In electromagnetism, flux is a scalar quantity, defined as the surface integral of the component of a vector field perpendicular to the surface at each point.
The electric field inside a dielectric decreases, when it is placed in an external electric field. This happens due to- a)Electrostatic shielding
- b)Polarization
- c)Electrostatic repulsion between atoms
- d)Electrostatic attraction between atoms
Correct answer is option 'B'. Can you explain this answer?
The electric field inside a dielectric decreases, when it is placed in an external electric field. This happens due to
a)
Electrostatic shielding
b)
Polarization
c)
Electrostatic repulsion between atoms
d)
Electrostatic attraction between atoms
|
Infinity Academy answered |
The external electric field polarizes the dielectric and an electric field is produced.
The net electric field inside the dielectric decreases due to polarization.
The net electric field inside the dielectric decreases due to polarization.
Three different capacitors are connected in series, then:- a)they will have equal charges
- b)they will have same potential
- c)both 1 & 2
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Three different capacitors are connected in series, then:
a)
they will have equal charges
b)
they will have same potential
c)
both 1 & 2
d)
none of these
|
Mira Joshi answered |
C = Q/V
series connection splits the battery potential, hence....
V = V1 + V2
V1 = Q/C1 & V2 = Q/C2
Therefore, both have equal charge
series connection splits the battery potential, hence....
V = V1 + V2
V1 = Q/C1 & V2 = Q/C2
Therefore, both have equal charge
- a)48 × 10 8 N
- b)4.8 × 10 9 N
- c)48 × 10 4 N
- d)4.8 × 10 10 N
Correct answer is 'D'. Can you explain this answer?
|
Ritika Sengupta answered |
What should be the radius of an isolated spherical conductor so that it has a capacity of 2μF?- a)1.8X105 m
- b)1.8X104 m
- c)2.5X104 m
- d)2.5X105 m
Correct answer is option 'B'. Can you explain this answer?
What should be the radius of an isolated spherical conductor so that it has a capacity of 2μF?
a)
1.8X105 m
b)
1.8X104 m
c)
2.5X104 m
d)
2.5X105 m
|
Gunjan Lakhani answered |
Capacitance of an isolated spherical conductor =4π€R , where €=permittivity of vacuum
C=2×10-6F
C=4π€R
4×3.14×8.85×10-12×R=2×10-6
R=2×10-6/4×3.14×8.85×10-12
=1.8×104 m
C=2×10-6F
C=4π€R
4×3.14×8.85×10-12×R=2×10-6
R=2×10-6/4×3.14×8.85×10-12
=1.8×104 m
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
Can you explain the answer of this question below:A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is- A:0
- B:
- C:
- D:
The answer is b.
A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is
A:
0
B:
C:
D:
|
Lavanya Menon answered |
Given that, A charge q is placed at the centre of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0
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.
If the electric field is given by
; calculate the electric flux through as surface of area 10 units lying in y-z plane.- a)50 units
- b)40 units
- c)60 units
- d)30 units
Correct answer is option 'C'. Can you explain this answer?
If the electric field is given by; calculate the electric flux through as surface of area 10 units lying in y-z plane.
; calculate the electric flux through as surface of area 10 units lying in y-z plane.a)
50 units
b)
40 units
c)
60 units
d)
30 units
|
EduRev JEE answered |
As surface of area lies in the Y-Z plane, thus its area vector points in X direction i.e. A=10i
Electric flux ϕ=E.A
∴ ϕ=(6i+3j+4k).(10i)
⟹ ϕ=6×10=60 unit
Electric flux ϕ=E.A
∴ ϕ=(6i+3j+4k).(10i)
⟹ ϕ=6×10=60 unit
- a)0.5 J
- b)1.12 J
- c)1.2 J
- d)0.9 J
Correct answer is 'D'. Can you explain this answer?
|
|
Nikita Singh answered |
q=6x10-10 c
Q=-2x10-9c
r1=3x10-2m
r2=4x10-2m
△vQ=w/q
(1/4πεo)-2x10-9/10-2[(1/4)-(1/3)]=W/6x10-3
9x109x2x10-9x6x10-3/12x10-2=W
W=9x10-3x102
W=0.9J
Q=-2x10-9c
r1=3x10-2m
r2=4x10-2m
△vQ=w/q
(1/4πεo)-2x10-9/10-2[(1/4)-(1/3)]=W/6x10-3
9x109x2x10-9x6x10-3/12x10-2=W
W=9x10-3x102
W=0.9J
The force between two small charged spheres having charges 3 x 10-6C and 4 x 10-6C placed 40 cm apart in air is- a)67.5 x 10-3 N
- b)67.5 x 10-2 N
- c)6.75 x 10-1 N
- d)none
Correct answer is option 'C'. Can you explain this answer?
The force between two small charged spheres having charges 3 x 10-6C and 4 x 10-6C placed 40 cm apart in air is
a)
67.5 x 10-3 N
b)
67.5 x 10-2 N
c)
6.75 x 10-1 N
d)
none
|
|
Siddharth Tiwari answered |
**Explanation:**
When two charges are placed in close proximity to each other, they exert a force on each other. This force is known as the electrostatic force. The electrostatic force between two charges is governed by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this case, we have a positive charge of 1C and a negative charge of -4C. According to Coulomb's Law, the force between these two charges will be attractive because one charge is positive and the other is negative. The force will act along the line connecting the two charges, from the positive charge to the negative charge.
To determine the direction of the force, we can use a simple mnemonic: "like charges repel, opposite charges attract." Since the charges in this scenario are opposite (positive and negative), they will attract each other. Therefore, the force will act from the positive charge (1C) towards the negative charge (-4C).
Hence, the correct answer is option C: "From 1C to -4C."
When two charges are placed in close proximity to each other, they exert a force on each other. This force is known as the electrostatic force. The electrostatic force between two charges is governed by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this case, we have a positive charge of 1C and a negative charge of -4C. According to Coulomb's Law, the force between these two charges will be attractive because one charge is positive and the other is negative. The force will act along the line connecting the two charges, from the positive charge to the negative charge.
To determine the direction of the force, we can use a simple mnemonic: "like charges repel, opposite charges attract." Since the charges in this scenario are opposite (positive and negative), they will attract each other. Therefore, the force will act from the positive charge (1C) towards the negative charge (-4C).
Hence, the correct answer is option C: "From 1C to -4C."
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.
Two equipotential surfaces have a potential of -10V and 90V respectively, what is the difference in potential between these surfaces?- a)90V
- b)80V
- c)0V
- d)100V
Correct answer is 'D'. Can you explain this answer?
Two equipotential surfaces have a potential of -10V and 90V respectively, what is the difference in potential between these surfaces?
a)
90V
b)
80V
c)
0V
d)
100V
|
|
Krishna Iyer answered |
Potential = high potential - low potential
so our high potential is 90V and low potential is 10V,
potential = 90-(-10)=100V
so our high potential is 90V and low potential is 10V,
potential = 90-(-10)=100V
The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is- a)Zero
- b)Infinity
- c)One
- d)Two
Correct answer is option 'A'. Can you explain this answer?
The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is
a)
Zero
b)
Infinity
c)
One
d)
Two
|
Om Desai answered |
Since Potential difference between two points in equipotential surfaces is zero, the work done between two points in equipotential surface is also zero.
Which of the following is a non polar dielectric?- a)Water
- b)Alcohol
- c)HCl
- d)Benzene
Correct answer is option 'D'. Can you explain this answer?
Which of the following is a non polar dielectric?
a)
Water
b)
Alcohol
c)
HCl
d)
Benzene
|
|
Riya Banerjee answered |
Alcohol and HCl are polar molecules since they have a net dipole moment towards a particular direction. Both water and benzene are non-polar molecules. But water is a conductor of electricity, whereas benzene is a dielectric (insulator).
1 microfarad is equal to- a)10-6F
- b)10-12F
- c)10-15F
- d)10-9F
Correct answer is option 'A'. Can you explain this answer?
1 microfarad is equal to
a)
10-6F
b)
10-12F
c)
10-15F
d)
10-9F
|
|
Lavanya Menon answered |
The microfarad (symbolized µF) is a unit of capacitance, equivalent to 0.000001 (10 to the -6th power) farad.
What will be the effect on capacitance, if the distance between the parallel plates reduced to one-third of its original value?- a)decrease by a factor 3
- b)increase by a factor 3
- c)increase by a factor 3/2
- d)increase by a factor ½
Correct answer is option 'B'. Can you explain this answer?
What will be the effect on capacitance, if the distance between the parallel plates reduced to one-third of its original value?
a)
decrease by a factor 3
b)
increase by a factor 3
c)
increase by a factor 3/2
d)
increase by a factor ½
|
Preeti Khanna answered |
C=(Kε0A/d)∝K/d
Hence,C1/C2=d2/d1=(d/3)d=3
C2=3C1
Hence,C1/C2=d2/d1=(d/3)d=3
C2=3C1
Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7 C? The radii of A and B are negligible compared to the distance of separation.- a)4.5×10−2N
- b)2.5×10−2 N
- c)1.5×10−2 N
- d)3.5×10−2 N
Correct answer is option 'C'. Can you explain this answer?
Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7 C? The radii of A and B are negligible compared to the distance of separation.
a)
4.5×10−2N
b)
2.5×10−2 N
c)
1.5×10−2 N
d)
3.5×10−2 N
|
|
Om Desai answered |
Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is- a)6.67J
- b)60J
- c)Zero
- d)15J
Correct answer is option 'C'. Can you explain this answer?
Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is
a)
6.67J
b)
60J
c)
Zero
d)
15J
|
|
Suresh Iyer answered |
The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q0=0.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q0=0.
Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC-1 and the electric potential at a point ‘B’ due to same charge is 6 JC-1. The distance between AB is- a)2 m
- b)1.5 m
- c)1 m
- d)0.5 m
Correct answer is option 'D'. Can you explain this answer?
Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC-1 and the electric potential at a point ‘B’ due to same charge is 6 JC-1. The distance between AB is
a)
2 m
b)
1.5 m
c)
1 m
d)
0.5 m
|
|
Nandini Patel answered |
As we know , E . l = V where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B ,
l = ( 6 / 12 ) m or ( 1 / 2 ) m = 0.5 m
Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1: Q2: Q3, is - a)it is 1:03:05
- b)it is 1:02:03
- c)it is 1:04:09
- d)it is 1:08:18
Correct answer is 'A'. Can you explain this answer?
Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1: Q2: Q3, is
a)
it is 1:03:05
b)
it is 1:02:03
c)
it is 1:04:09
d)
it is 1:08:18
|
|
Sanaya Kumar answered |
Since the charge goes to outer surface when given inside so the charges on concentri shells are respectively.
Q1, Q1 + Q2, Q1 + Q2 + Q3. Sine their charge densities are equal so σ1 = σ2 = σ3
So Q1 = 3Q2 = 5Q3 so 1 : 3 : 5
The polarized dielectric is equivalent to- a)non-zero volume charge density
- b)one charged surface with induced surface charge density
- c)two charged surfaces with induced surface charge densities
- d)zero surface charge density
Correct answer is option 'C'. Can you explain this answer?
The polarized dielectric is equivalent to
a)
non-zero volume charge density
b)
one charged surface with induced surface charge density
c)
two charged surfaces with induced surface charge densities
d)
zero surface charge density
|
Learners Habitat answered |
A polarized dielectric is equivalent to two charged surfaces with induced charges (polarization charges).
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
As shown in the figure below, an ellipsoidal cavity is carved within a perfect conductor. A positive charge Q is placed at the centre of the cavity. If points A and b are shown on the cavity surface (see figure), then which among the following choices is correct?
- a)Potential at A = Potential at B
- b)Charge density at A = Charge density at B
- c)Electric field near A = Electric field near B
- d)Total electric flux at the surface of the cavity = zero
Correct answer is option 'A'. Can you explain this answer?
As shown in the figure below, an ellipsoidal cavity is carved within a perfect conductor. A positive charge Q is placed at the centre of the cavity. If points A and b are shown on the cavity surface (see figure), then which among the following choices is correct?
a)
Potential at A = Potential at B
b)
Charge density at A = Charge density at B
c)
Electric field near A = Electric field near B
d)
Total electric flux at the surface of the cavity = zero
|
|
Om Desai answered |
Electric field at A is different from field at B because, E= K/r2
we know that conductor is an equipotential surface, so potential will be the same at A and B.
As charge density, σ∝1/r, then charge density is different at A and B.
we know that conductor is an equipotential surface, so potential will be the same at A and B.
As charge density, σ∝1/r, then charge density is different at A and B.
Electric flux density is a function of _______.
- a)Volume
- b)Charge
- c)Current
- d)Voltage
Correct answer is option 'B'. Can you explain this answer?
Electric flux density is a function of _______.
a)
Volume
b)
Charge
c)
Current
d)
Voltage
|
|
Aarya Khanna answered |
Electric flux density is a measure of the electric flux passing through a given area. It is defined as the amount of electric flux passing through a unit area perpendicular to the electric field. The SI unit of electric flux density is coulombs per square meter (C/m²).
Factors affecting electric flux density:
Charge:
Electric flux density is directly proportional to the amount of charge present in a given area. The higher the charge density, the higher the electric flux density.
Area:
Electric flux density is inversely proportional to the area of the surface through which the flux passes. The smaller the area, the higher the electric flux density.
Distance:
Electric flux density is inversely proportional to the distance of the charge from the surface through which the flux passes. The closer the charge, the higher the electric flux density.
Dielectric constant:
Electric flux density is affected by the dielectric constant of the material through which the flux passes. The higher the dielectric constant, the higher the electric flux density.
Conclusion:
From the above explanation, it is clear that electric flux density is a function of charge.
Factors affecting electric flux density:
Charge:
Electric flux density is directly proportional to the amount of charge present in a given area. The higher the charge density, the higher the electric flux density.
Area:
Electric flux density is inversely proportional to the area of the surface through which the flux passes. The smaller the area, the higher the electric flux density.
Distance:
Electric flux density is inversely proportional to the distance of the charge from the surface through which the flux passes. The closer the charge, the higher the electric flux density.
Dielectric constant:
Electric flux density is affected by the dielectric constant of the material through which the flux passes. The higher the dielectric constant, the higher the electric flux density.
Conclusion:
From the above explanation, it is clear that electric flux density is a function of charge.
Which of the following molecule has permanent dipole moment?- a)HCl
- b)CO2
- c)CH4
- d)BF3
Correct answer is option 'A'. Can you explain this answer?
Which of the following molecule has permanent dipole moment?
a)
HCl
b)
CO2
c)
CH4
d)
BF3
|
Manoj Chauhan answered |
Explanation:
A permanent dipole moment is created when there is an electronegativity difference between two non-identical atoms, resulting in an uneven distribution of charge. The molecule acquires a positive and negative end, creating a dipole moment.
HCl molecule has a permanent dipole moment because of the difference in electronegativity between H and Cl atoms. Chlorine has a higher electronegativity than hydrogen. Therefore, the electrons in the covalent bond are drawn closer to the chlorine atom, resulting in a partial negative charge on the chlorine atom and a partial positive charge on the hydrogen atom. Thus, HCl has a dipole moment.
On the other hand, CO2, CH4, and BF3 do not have a permanent dipole moment because they are symmetrical molecules where the electronegativity of the atoms is the same. In CO2, the electronegativity of the carbon and oxygen atoms is the same, so the molecule has no dipole moment. In CH4, the four hydrogen atoms are arranged symmetrically around the carbon atom, resulting in no net dipole moment. Similarly, in BF3, the three fluorine atoms are arranged symmetrically around the boron atom, resulting in no net dipole moment.
Therefore, the correct option is A, HCl, which has a permanent dipole moment due to the difference in electronegativity between H and Cl atoms.
A permanent dipole moment is created when there is an electronegativity difference between two non-identical atoms, resulting in an uneven distribution of charge. The molecule acquires a positive and negative end, creating a dipole moment.
HCl molecule has a permanent dipole moment because of the difference in electronegativity between H and Cl atoms. Chlorine has a higher electronegativity than hydrogen. Therefore, the electrons in the covalent bond are drawn closer to the chlorine atom, resulting in a partial negative charge on the chlorine atom and a partial positive charge on the hydrogen atom. Thus, HCl has a dipole moment.
On the other hand, CO2, CH4, and BF3 do not have a permanent dipole moment because they are symmetrical molecules where the electronegativity of the atoms is the same. In CO2, the electronegativity of the carbon and oxygen atoms is the same, so the molecule has no dipole moment. In CH4, the four hydrogen atoms are arranged symmetrically around the carbon atom, resulting in no net dipole moment. Similarly, in BF3, the three fluorine atoms are arranged symmetrically around the boron atom, resulting in no net dipole moment.
Therefore, the correct option is A, HCl, which has a permanent dipole moment due to the difference in electronegativity between H and Cl atoms.
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.
If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is -5 V to another place, where potential is V volt. The value of V is- a)15 V
- b)20 V
- c)25 V
- d)10 V
Correct answer is option 'B'. Can you explain this answer?
If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is -5 V to another place, where potential is V volt. The value of V is
a)
15 V
b)
20 V
c)
25 V
d)
10 V
|
|
Suresh Iyer answered |
From the definition, the work done to a test charge ‘q0’ from one place to another place in an electric field is given by the formula
W=q0x[vfinal-vinitial ]
100=4x[v-(-5)]
v+5=25
v=20V
W=q0x[vfinal-vinitial ]
100=4x[v-(-5)]
v+5=25
v=20V
Can you explain the answer of this question below:Equal charges are given to two spheres of different radii. The potential will
- A:
be equal on both the spheres
- B:
be more on the smaller sphere
- C:
be more on the bigger sphere
- D:
depend on the material of the sphere
The answer is b.
Equal charges are given to two spheres of different radii. The potential will
be equal on both the spheres
be more on the smaller sphere
be more on the bigger sphere
depend on the material of the sphere
|
Ruchika Agarwal answered |
V(potential) is inversely proportional to r ( radius).so smaller sphere will have more potential as its radius will be small.
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface cantered at the origin with r = R- a)2Rλ/ε0
- b)Rλ/ε0
- c)λ/ε0
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface cantered at the origin with r = R
a)
2Rλ/ε0
b)
Rλ/ε0
c)
λ/ε0
d)
none of the above
|
Neha Chauhan answered |
Total charge inside the spherical surface =2Rλ
∫E.dS = qin/ε0
So flux = 2Rλ/ε0
Hence answer is (a)
∫E.dS = qin/ε0
So flux = 2Rλ/ε0
Hence answer is (a)
The electrostatic potential energy between two charges q1 and q2 separated by a distance by r is given by- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The electrostatic potential energy between two charges q1 and q2 separated by a distance by r is given by
a)
b)
c)
d)
|
Khushi Mittal answered |
Energy = force × distance U=Kq1q2/r^2 × r U=Kq1q2/r
A hollow metal sphere of radius 5cm is charged so that the potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is- a)4V
- b)zero
- c)10/3V
- d)10V
Correct answer is option 'D'. Can you explain this answer?
A hollow metal sphere of radius 5cm is charged so that the potential on its surface is 10V. The potential at a distance of 2cm from the centre of the sphere is
a)
4V
b)
zero
c)
10/3V
d)
10V
|
Ayush Joshi answered |
In the case of a hollow metal sphere (spherical shell), the electric field inside the shell is zero. This means that the potential inside the shell is constant. Therefore the potential at the centre of the sphere is the same as that on its surface, i.e. 10 V.
The diagram shows the electric field lines in a region of space containing two small charged spheres, Y and Z then which statement is true?
- a)Magnitude of electric field is same everywhere
- b)Y is negative & Z is positive
- c)Y and Z must have the same sign
- d)A small negatively charged body placed at X would be pushed to the right
Correct answer is option 'D'. Can you explain this answer?
The diagram shows the electric field lines in a region of space containing two small charged spheres, Y and Z then which statement is true?
a)
Magnitude of electric field is same everywhere
b)
Y is negative & Z is positive
c)
Y and Z must have the same sign
d)
A small negatively charged body placed at X would be pushed to the right
|
Divey Sethi answered |
A small negatively charged body placed at X would be pushed to the right
The negative charge feels a force opposite to the direction of field.
Other cases are not possible from properties of electric lines of force.
Can you explain the answer of this question below:At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?
- A:
at the surface of spherical shell
- B:
outside the spherical shell
- C:
inside the spherical shell
- D:
at the centre of spherical shell
The answer is a.
At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?
at the surface of spherical shell
outside the spherical shell
inside the spherical shell
at the centre of spherical shell
|
Bittu Raj Bittu answered |
This is because of when a charge given to a hollow sphere that is spherical shell then all charges will reside on its surface and as we know that electric field is directly proportional to given charge . so according to this concept we can say about this point.
Can you explain the answer of this question below:The electric field intensity due to a sphere (solid or hollow) at an external point varies as-
- A:
1/r
- B:
does not depend upon r
- C:
1/r3
- D:
1/r2
The answer is d.
The electric field intensity due to a sphere (solid or hollow) at an external point varies as-
1/r
does not depend upon r
1/r3
1/r2
|
|
Agnal Jose answered |
It followw inverse square law as distance of observational point from source charge increases the electric field intensity decreases hence option d is correct
Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.- a)480 C
- b)478 C
- c)450 C
- d)500 C
Correct answer is option 'A'. Can you explain this answer?
Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.
a)
480 C
b)
478 C
c)
450 C
d)
500 C
|
|
Nikita Singh answered |
C1= 20×10µf
and C2= 30×10µf
in series Ceq = C1C2/(C1+C2)
Ceq = 20×10^(-6)×30×10^(-6)/20×10^(-6)+30^×10(-6)
Ceq= 12×10^(-6)f
As we know that Q = CV
Putting the values of C and V= 40V, we get
Q = (12 * 10^-6) * 40
= 480µC
Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.
- a)6 μƒ
- b)3 μƒ
- c)4 μƒ
- d)2 μƒ
Correct answer is option 'B'. Can you explain this answer?
Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.
a)
6 μƒ
b)
3 μƒ
c)
4 μƒ
d)
2 μƒ
|
|
Lavanya Menon answered |
Combination of Capacitors in Series
C(effective) = C/3 = 1
C(effective) = C/3 = 1
When two capacitors C1 and C2 are connected in series and parallel, their equivalent capacitances comes out to be 3 μƒ and 16 μƒ respectively. Calculate values of C1 and C2.- a)4μƒ and 12μƒ
- b)6μƒ and 10μƒ
- c)4μƒ and 10μƒ
- d)6μƒ and 12μƒ
Correct answer is option 'A'. Can you explain this answer?
When two capacitors C1 and C2 are connected in series and parallel, their equivalent capacitances comes out to be 3 μƒ and 16 μƒ respectively. Calculate values of C1 and C2.
a)
4μƒ and 12μƒ
b)
6μƒ and 10μƒ
c)
4μƒ and 10μƒ
d)
6μƒ and 12μƒ
|
Anonymous answered |
Electric field of a point charge extends up to r where 1/r is- a)one
- b)infinity
- c)Zero
- d)two
Correct answer is option 'C'. Can you explain this answer?
Electric field of a point charge extends up to r where 1/r is
a)
one
b)
infinity
c)
Zero
d)
two
|
Sherlin Dsouza answered |
Hii.... so electric extends upto infinity so inverse of infinity is zero... regards SHERLIN
A neutral point lies at the centre of the line joining the charges when- a)Two charges are unequal and opposite
- b)Two charges are equal and of same sign
- c)Two charges are unequal and of same sign
- d)Two charges are equal and of opposite sign
Correct answer is option 'B'. Can you explain this answer?
A neutral point lies at the centre of the line joining the charges when
a)
Two charges are unequal and opposite
b)
Two charges are equal and of same sign
c)
Two charges are unequal and of same sign
d)
Two charges are equal and of opposite sign
|
|
Shreya Gupta answered |
Neutral point due to a system of two like point charges For this case neutral point is obtained at an internal point along the line joining two like charges.Here neutral point lies outside the line joining two unlike charges and also it lies nearer to the charge which is smaller in magnitude.
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