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All questions of Electrostatic Potential and Capacitance for JEE Exam
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 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
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
|
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
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).
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.
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
There are three capacitors with equal capacitance. In series combination, they have a net capacitance of C1 and in parallel combination, a net capacitance of C2.What will be the value of C1 C2?- a)1:9
- b)3:2
- c)2:3
- d)9:1
Correct answer is option 'A'. Can you explain this answer?
There are three capacitors with equal capacitance. In series combination, they have a net capacitance of C1 and in parallel combination, a net capacitance of C2.What will be the value of C1 C2?
a)
1:9
b)
3:2
c)
2:3
d)
9:1
|
Dilip Chaurasiya answered |
In series C= C/3 and in parallel C= 3C divide it we get 1/9
Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between the points M and N. 
- a)14
- b)12
- c)10
- d)15
Correct answer is option 'B'. Can you explain this answer?
Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between the points M and N.
a)
14
b)
12
c)
10
d)
15
|
|
Krishna Iyer answered |
Five capacitors 4μF, 2μF, 4μF, 6μF and 1.2μF are connected as shown in the figure.
Here, Equivalent capacitance = ½,
EMF = 24
Thus, charge passes through each capacitor = 12 μC
Voltage across 4μF = 3V
Voltage across 6μF = 2V
=> Potential difference (V):
V = 3 + 7+ 2 = 12V
Thus, potential difference (V) between the two given points M and N is 12v.
Here, Equivalent capacitance = ½,
EMF = 24
Thus, charge passes through each capacitor = 12 μC
Voltage across 4μF = 3V
Voltage across 6μF = 2V
=> Potential difference (V):
V = 3 + 7+ 2 = 12V
Thus, potential difference (V) between the two given points M and N is 12v.
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.
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
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.
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.
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.
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
|
Ashish Bhoumick answered |
Because charge distribute surface of the area.here small sphere has high density charge.so the answer 'b' is right...
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
Electric potential at a point located far away from the charge is taken to be- a)Unity
- b)May be zero or infinite
- c)Zero
- d)Infinite
Correct answer is option 'C'. Can you explain this answer?
Electric potential at a point located far away from the charge is taken to be
a)
Unity
b)
May be zero or infinite
c)
Zero
d)
Infinite
|
Suman Kumar answered |
Because the distance is much larger so electric potential is 0 in other words there is no effect of other charge on that charge
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).
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
Three uncharged capacitors of capacitane C1 = 1mF, C2 = 2mF and C3 = 3mF are connected as shown in figure to one another and to points A, B and D potential fA = 10V, fB = 25V and fD = 20 V, Determine the potential (f0) at point O.
- a)20 V
- b)30 V
- c)40 V
- d)10 V
Correct answer is option 'A'. Can you explain this answer?
Three uncharged capacitors of capacitane C1 = 1mF, C2 = 2mF and C3 = 3mF are connected as shown in figure to one another and to points A, B and D potential fA = 10V, fB = 25V and fD = 20 V, Determine the potential (f0) at point O.
a)
20 V
b)
30 V
c)
40 V
d)
10 V
|
|
Lavanya Menon answered |
⇒q1+q2+q3=0
⇒C1(Vo−VA)+C2(Vo−VB)+C3(Vo−VD)=0
⇒(C1+C2+C3)Vo=C1VA+C2VB+C3VD
Vo= C1VA+C2VB+C3VD/ C1+C2+C3
= [(1×10)+(2×25)+(3×20)]/(1+2+3)
=(10+50+60)/6
=20Volts
⇒C1(Vo−VA)+C2(Vo−VB)+C3(Vo−VD)=0
⇒(C1+C2+C3)Vo=C1VA+C2VB+C3VD
Vo= C1VA+C2VB+C3VD/ C1+C2+C3
= [(1×10)+(2×25)+(3×20)]/(1+2+3)
=(10+50+60)/6
=20Volts
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 volt. The potential at the centre of the sphere is- a)8 volt
- b)zero
- c)800 volt
- d)80 volt
Correct answer is option 'D'. Can you explain this answer?
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 volt. The potential at the centre of the sphere is
a)
8 volt
b)
zero
c)
800 volt
d)
80 volt
|
Sunidhi Pandey answered |
Chage is uniformly distributed in an hollow sphere.
μμf is also called?- a)nanofarad
- b)microfarad
- c)picofarad
- d)fermifarad
Correct answer is option 'C'. Can you explain this answer?
μμf is also called?
a)
nanofarad
b)
microfarad
c)
picofarad
d)
fermifarad
|
|
Arghya answered |
10^(-6)=microfarad.... so ,10^(-6)*10^(-6)=10^(-12)=PICOFARAD
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 |
When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy- a)Remains constant
- b)Decreases
- c)Increases
- d)Either increase or remain constant
Correct answer is option 'C'. Can you explain this answer?
When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy
a)
Remains constant
b)
Decreases
c)
Increases
d)
Either increase or remain constant
|
Siddhi Bhardwaj answered |
You can generalise it by assuming a positive charge moving away from another positive charge. now both of them are repelling each other with some force. so that positive charge will accelerate which results in the increase in K.E.
A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearly- a)7 × 10_4 J
- b)3.5 × 10_4 J
- c)14 × 10_4 J
- d)7 × 10_3 J
Correct answer is option 'A'. Can you explain this answer?
A conductor of capacitance 0.5mF has been charged to 100volts. It is now connected to uncharged conductor of capacitance 0.2mF. The loss in potential energy is nearly
a)
7 × 10_4 J
b)
3.5 × 10_4 J
c)
14 × 10_4 J
d)
7 × 10_3 J
|
Jyoti Sengupta answered |
Given,
C1=0.5 mf
C2=0.2mf
V=100v
So,
U=(1/2) CTV2
CT=C1C2/XC1+C2=(0.5x0.2/0.5+0.2)mf
CT=0.14mf
U=(1/2)(0.14mf)(100V)2
=(10.07x1-6x104)J
=(7x10-2x10-6x104)J
=7x10-4J
C1=0.5 mf
C2=0.2mf
V=100v
So,
U=(1/2) CTV2
CT=C1C2/XC1+C2=(0.5x0.2/0.5+0.2)mf
CT=0.14mf
U=(1/2)(0.14mf)(100V)2
=(10.07x1-6x104)J
=(7x10-2x10-6x104)J
=7x10-4J
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
The shape of equipotential surface for an infinite line charge is:- a)Coaxial cylindrical surfaces
- b)Parallel plane surfaces
- c)Parallel plane surfaces perpendicular to lines of force
- d)None of above
Correct answer is option 'A'. Can you explain this answer?
The shape of equipotential surface for an infinite line charge is:
a)
Coaxial cylindrical surfaces
b)
Parallel plane surfaces
c)
Parallel plane surfaces perpendicular to lines of force
d)
None of above
|
|
Rajesh Gupta answered |
The shape of equipotential surface for an infinite line charge is coaxial cylindrical because A curved surface on which potential is constant is equipotential curve . If we consider the line charge then the focus of the point should have the same potential hence it is a coaxial cylinder.
The distance between the plates of a capacitor is d. What will be the new capacitance if a metal plate of thickness d/2 is introduced between the plates without touching them- a)it will be thrice of its initial value
- b)it will be double of its initial value
- c)remains the same
- d)it will be half of its initial value
Correct answer is option 'B'. Can you explain this answer?
The distance between the plates of a capacitor is d. What will be the new capacitance if a metal plate of thickness d/2 is introduced between the plates without touching them
a)
it will be thrice of its initial value
b)
it will be double of its initial value
c)
remains the same
d)
it will be half of its initial value
|
|
Nisha Patel answered |
Explanation:
Capacitance of a capacitor:
The capacitance of a capacitor is given by the equation:
C = εA/d
where C is the capacitance, ε is the permittivity of the medium between the plates, A is the area of the plates, and d is the distance between the plates.
Effect of introducing a metal plate:
When a metal plate of thickness d/2 is introduced between the plates without touching them, the distance between the plates decreases by d/2. This means that the new distance between the plates is d/2.
New capacitance:
Using the equation for capacitance, we can calculate the new capacitance as:
C' = εA/(d/2)
C' = 2εA/d
The new capacitance is double the initial capacitance. Therefore, the correct answer is option B, "it will be double of its initial value."
Capacitance of a capacitor:
The capacitance of a capacitor is given by the equation:
C = εA/d
where C is the capacitance, ε is the permittivity of the medium between the plates, A is the area of the plates, and d is the distance between the plates.
Effect of introducing a metal plate:
When a metal plate of thickness d/2 is introduced between the plates without touching them, the distance between the plates decreases by d/2. This means that the new distance between the plates is d/2.
New capacitance:
Using the equation for capacitance, we can calculate the new capacitance as:
C' = εA/(d/2)
C' = 2εA/d
The new capacitance is double the initial capacitance. Therefore, the correct answer is option B, "it will be double of its initial value."
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
Can you explain the answer of this question below:In a parallel plate capacitor the potential difference of 150 V is maintained between the plates. If distance between the plates is 6 mm, what will be the electric field at points A and B?
- A:
2.5×103
- B:
2.5×104
- C:
2.5×105
- D:
3.5×104
The answer is b.
In a parallel plate capacitor the potential difference of 150 V is maintained between the plates. If distance between the plates is 6 mm, what will be the electric field at points A and B?
2.5×103
2.5×104
2.5×105
3.5×104
|
|
Om Desai answered |
Calculate the equivalent capacitance for the following combination between points A and B
- a)12/5 μƒ
- b)5 μƒ
- c)40/9 μƒ
- d)37/5 μƒ
Correct answer is option 'D'. Can you explain this answer?
Calculate the equivalent capacitance for the following combination between points A and B
a)
12/5 μƒ
b)
5 μƒ
c)
40/9 μƒ
d)
37/5 μƒ
|
Divey Sethi answered |
4μf and 6μf are in series (right side of AB) so 6x(4/6)+4=24/10μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf
A particle of charge q1 = 3μC is located on x-axis at the point x1 = 6 cm. A second point charge q2 = 2μC is placed on the x-axis at x2 = -4 cm. The absolute electric potential at the origin is- a)9 x 10-6 V
- b)9 x 105 V
- c)9 x 10-4 V
- d)9 x 10-2 V
Correct answer is option 'C'. Can you explain this answer?
A particle of charge q1 = 3μC is located on x-axis at the point x1 = 6 cm. A second point charge q2 = 2μC is placed on the x-axis at x2 = -4 cm. The absolute electric potential at the origin is
a)
9 x 10-6 V
b)
9 x 105 V
c)
9 x 10-4 V
d)
9 x 10-2 V
|
|
Nikita Singh answered |
V=kq/r
Therefore, Vnet=Kx[(2µc/4) + (3µc/6)]
=9c109x10-6[1/2+1/2]x102
=9x105
Therefore, Vnet=Kx[(2µc/4) + (3µc/6)]
=9c109x10-6[1/2+1/2]x102
=9x105
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1, k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by
- a)
- b)
- c)
- d)K = K1 + K2 + 2K3
Correct answer is option 'B'. Can you explain this answer?
A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1, k2 and k3 as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by
a)
b)
c)
d)
K = K1 + K2 + 2K3
|
Bs Academy answered |
Let C1 = Capacity of capacitor with K1
C2 = Capacity of capacitor with K2
C3 = Capacity of capacitor with K3
C2 = Capacity of capacitor with K2
C3 = Capacity of capacitor with K3
C1 and C2 are in parallel
Ceq and C3 are in series
But 
for single equivalent capacitor
for single equivalent capacitor
Electric field at the surface of a charged conductor is proportional to- a)Surface charge density
- b)Volume of the sphere
- c)Volume charge density
- d)Area of the sphere
Correct answer is option 'A'. Can you explain this answer?
Electric field at the surface of a charged conductor is proportional to
a)
Surface charge density
b)
Volume of the sphere
c)
Volume charge density
d)
Area of the sphere
|
Ciel Knowledge answered |
Electric field at the surfaces of charged conductors is σ/ε0.n, where n is a unit vector normal to the surface.
We clearly see that the electric field is perpendicular to surface charge density (σ).
We clearly see that the electric field is perpendicular to surface charge density (σ).
Electric potential is- a)scalar and dimensionless
- b)vector and dimensionless
- c)scalar with dimension
- d)vector with dimension
Correct answer is option 'C'. Can you explain this answer?
Electric potential is
a)
scalar and dimensionless
b)
vector and dimensionless
c)
scalar with dimension
d)
vector with dimension
|
|
Pooja Mehta answered |
He electric potential due to a system of point charges is equal to the sum of the point charges' individual potentials. This fact simplifies calculations significantly, since addition of potential (scalar) fields is much easier than addition of the electric (vector) fields.
Can you explain the answer of this question below:The potential energy of a system containing only one point charge is
- A:
Zero
- B:
Infinity
- C:
Nonzero finite
- D:
None of the above
The answer is a.
The potential energy of a system containing only one point charge is
Zero
Infinity
Nonzero finite
None of the above
|
.mie. answered |
Answer is 0 as there are no other sources of electrostatic potential .... against which an external agent must do work.... in moving the point charge.... from infinity to its final location.... therefore correct opt is A
The dipole moment per unit volume is called- a)Polarization
- b)Surface charge density
- c)Linear charge density
- d)Volume charge density
Correct answer is option 'A'. Can you explain this answer?
The dipole moment per unit volume is called
a)
Polarization
b)
Surface charge density
c)
Linear charge density
d)
Volume charge density
|
Preeti Iyer answered |
The electric dipole moment per unit volume is called polarization or polarization density (vector p).
It is always directed from negative charge to positive charge.
If there are N atoms per unit volume than
vector p = N (vector p)
where p is the electric dipole moment of individual atom.
In bringing an electron towards another electron, the electrostatic potential energy of the system- a)becomes zero
- b)decreases
- c)remains same
- d)increases
Correct answer is option 'D'. Can you explain this answer?
In bringing an electron towards another electron, the electrostatic potential energy of the system
a)
becomes zero
b)
decreases
c)
remains same
d)
increases
|
|
Nandini Patel answered |
The electron has negative charge. When an electron is bringing towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative: Electrostatic potential energy of system of two electrons
U=[1/4πε0][(−e)(−e)/r] = [1/4πε0](e^2/r)
Thus, as r decreases, potential energy U increases.
If a unit charge is taken from one part to another part over an equipotential surface, then what is the change in electrostatic potential energy of the charge?- a)10 J
- b)100 J
- c)1 J
- d)0 J
Correct answer is option 'D'. Can you explain this answer?
If a unit charge is taken from one part to another part over an equipotential surface, then what is the change in electrostatic potential energy of the charge?
a)
10 J
b)
100 J
c)
1 J
d)
0 J
|
Utsav Srivastava answered |
Equipotential surface means the potential on every. point on that surface is constant. it means the change in potential on equipotential surface is zero we know that... ( electrostatic potential energy = change in potential × charge.)... ... according to this electrostatic potential energy is zero
Dimensional formula for potential difference is- a)[ML2 T-2 A-1]
- b)[ML2 T-3 A-1]
- c)[ML T-1 A-1]
- d)[M2L T-2 A-1]
Correct answer is option 'B'. Can you explain this answer?
Dimensional formula for potential difference is
a)
[ML2 T-2 A-1]
b)
[ML2 T-3 A-1]
c)
[ML T-1 A-1]
d)
[M2L T-2 A-1]
|
Rohit Shah answered |
Potential Diff = Volts = V = Workdone/Charge(Q)
Workdone = [Mass][Length]^2/[Time]^2
Hence, Potential Diff = [Mass][Length]^2[Time]^−2[Charge(Q)]^−1
Hence, The correct answer is Option A.
Two capacitors of 2mF and 3mF are charged to 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5 mF falls to the free ends of the wire. Then 
- a) Charge on the 1.5 mF capacitor is 180 mC
- b)Charge on the 2 mF capacitor is 120 mC
- c)Charge flows through A from right to left
- d)Charge flows through A from left to right
Correct answer is option 'A,B,C'. Can you explain this answer?
Two capacitors of 2mF and 3mF are charged to 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5 mF falls to the free ends of the wire. Then
a)
Charge on the 1.5 mF capacitor is 180 mC
b)
Charge on the 2 mF capacitor is 120 mC
c)
Charge flows through A from right to left
d)
Charge flows through A from left to right
|
|
Neha Sharma answered |
Charge in 2μF Capacitor =300μC
Charge in 3μF Capacitor =360μC
when 1.5μF falls on the circuit ends, Redistribution of charge now will be:
Charge in 2μF Capacitor =300−xμC
Charge in 3μF Capacitor =360−xμC
Charge in 1.5μF Capacitor =xμC
By Kirchhoff Voltage Law:
x/1.5=[(300−x)/2]+[(360−x)/3]
which gives x=180μC
Therefore, Answers are A, B & C
Charge in 3μF Capacitor =360μC
when 1.5μF falls on the circuit ends, Redistribution of charge now will be:
Charge in 2μF Capacitor =300−xμC
Charge in 3μF Capacitor =360−xμC
Charge in 1.5μF Capacitor =xμC
By Kirchhoff Voltage Law:
x/1.5=[(300−x)/2]+[(360−x)/3]
which gives x=180μC
Therefore, Answers are A, B & C
If the potential difference between the plates of a capacitor is increased by 0.1%, the energy stored in the capacitor increases by very nearly- a)0.1%
- b)0.144%
- c)0.11%
- d)0.2%
Correct answer is option 'D'. Can you explain this answer?
If the potential difference between the plates of a capacitor is increased by 0.1%, the energy stored in the capacitor increases by very nearly
a)
0.1%
b)
0.144%
c)
0.11%
d)
0.2%
|
Aastha Kulshreshtha answered |
Energy stored in capacitor= (1/2)CV^2
As there is no change in capacity but small change in potential diff. then the change in energy stored is given as:
∆U/U = 2�∆V/V. (as power associated with potential diff was 2)
Therefore % change
∆U/U �100 = 2 � ∆V/V �100
Thus,
% change in U = 2� 0.1%
% change in U = 0.2%
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