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All questions of Electrostatic Potential & Capacitance for JAMB 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?

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
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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?

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

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?

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?

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).

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?

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.

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?

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
 

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?

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. 

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?

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 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?

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."

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?

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.

.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

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?

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).
 

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?

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. 

when two capacitors are put in series, the equivalent capacitance is 
  • a)
    the reciprocal of the capacitances
  • b)
    smaller than both capacitances
  • c)
    the sum of the capacitances
  • d)
    the product of the capacitances 
Correct answer is option 'B'. Can you explain this answer?

Sreemoyee Sengupta answered
Explanation:

When two capacitors are put in series, the equivalent capacitance is smaller than both capacitances.

Why is this so? Let's look at the equation for capacitance in series:

1/Ceq = 1/C1 + 1/C2

Where Ceq is the equivalent capacitance and C1 and C2 are the capacitances of the individual capacitors. From this equation, we can see that the sum of the reciprocals of the capacitances is equal to the reciprocal of the equivalent capacitance.

Now, let's consider what happens when we put two capacitors in series. In this case, the capacitors share the same charge, so the voltage across each capacitor is different. This means that the capacitances are not simply added together, but rather combine in a way that reduces the overall capacitance.

To see why this is the case, let's take a simple example. Suppose we have two capacitors, each with a capacitance of 1 microfarad. If we put these capacitors in series, the equivalent capacitance is:

1/Ceq = 1/1 + 1/1 = 1/2

Ceq = 2 microfarads

As we can see, the equivalent capacitance is smaller than the capacitance of either capacitor alone. This is because the capacitors effectively "cancel out" some of each other's capacitance due to the shared charge and differing voltages.

Overall, when two capacitors are put in series, the equivalent capacitance is smaller than both capacitances because the capacitors "cancel out" some of each other's capacitance.

What is the direction of the lines of force at any point on the equipotential surface?
  • a)
    Parallel to it
  • b)
    Perpendicular to it.
  • c)
    Inclined at 45 degrees
  • d)
    None of the above
Correct answer is option 'B'. Can you explain this answer?

Anjali Iyer answered
When electric lines of force get perpendicular to equipotential surface then area vector and electric lines of force are parallel to each other.so angle between them is zero. due to that reason they get perpendicular to each other.

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?

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

Find the band gap energy when a light of wavelength 1240nm is incident on it.
  • a)
    1eV
  • b)
    2eV
  • c)
    3eV
  • d)
    4eV
Correct answer is option 'A'. Can you explain this answer?

Khushi Singh answered
Find the band gap energy when a light of wavelength 1240nm is incident on it. Explanation: The band gap energy in electron volt when wavelength is given is, Eg = 1.24(μm)/λ = 1.24 x 10-6/1240 x 10-9 = 1eV.

The potential energy of a system containing only one point charge is
  • a)
    Zero
  • b)
    Infinity
  • c)
    Non zero finite
  • d)
    None of the above
Correct answer is option 'A'. Can you explain this answer?

Manoj Chauhan answered
Explanation:

Potential energy is defined as the work done by an external force in bringing a system from infinity to its position. Hence, the potential energy of a system containing only one point charge is given by -

U = qV

where, q is the charge of the point charge and V is the potential at its position.

Now, let's consider the two cases -

Case 1: When the point charge is at infinity

At infinity, the potential is zero as the electric field due to a point charge decreases as we move away from it. Hence, the potential energy of the system containing only one point charge at infinity is zero.

U = qV = q x 0 = 0

Case 2: When the point charge is at a finite distance from infinity

In this case, the potential energy of the system will be non-zero finite as the potential at the position of the point charge will be non-zero.

U = qV ≠ 0

Conclusion:

Hence, the correct answer is option 'A' i.e. zero as the potential energy of a system containing only one point charge is zero when the point charge is at infinity.

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?

Amar Shah answered
**Explanation:**

To understand why the correct answer is option 'D', let's consider the equation for the energy stored in a capacitor:

**Energy (E) = 1/2 * C * V^2**

where E is the energy, C is the capacitance, and V is the potential difference (voltage) across the capacitor plates.

According to the question, the potential difference between the plates of the capacitor is increased by 0.1%. Let the initial potential difference be V1 and the increased potential difference be V2.

**Change in Potential Difference = (V2 - V1) = 0.1% of V1**

Using this information, we can write the new potential difference as:

**V2 = V1 + (0.1/100) * V1 = V1 + 0.001 * V1 = (1 + 0.001) * V1 = 1.001 * V1**

Next, we need to find the change in energy stored in the capacitor. Let the initial energy be E1 and the increased energy be E2.

**Change in Energy = (E2 - E1) = ?**

Using the equation for energy, we can write the new energy as:

**E2 = 1/2 * C * V2^2 = 1/2 * C * (1.001 * V1)^2 = 1/2 * C * (1.001^2) * V1^2**

Now, let's calculate the change in energy:

**Change in Energy = (E2 - E1) = 1/2 * C * (1.001^2) * V1^2 - 1/2 * C * V1^2**

Simplifying this expression:

**Change in Energy = 1/2 * C * V1^2 * (1.001^2 - 1) = 1/2 * C * V1^2 * (1.001^2 - 1)**

To find the percentage increase in energy, we need to divide the change in energy by the initial energy and multiply by 100. Let's calculate:

**Percentage Increase in Energy = (Change in Energy / E1) * 100**

Substituting the expression for change in energy:

**Percentage Increase in Energy = [1/2 * C * V1^2 * (1.001^2 - 1)] / [1/2 * C * V1^2] * 100**

Simplifying this expression:

**Percentage Increase in Energy = (1.001^2 - 1) * 100 = (1.001 * 1.001 - 1) * 100 = (1.002001 - 1) * 100 = 0.2001 * 100 = 0.2%**

Therefore, the energy stored in the capacitor increases by approximately 0.2% when the potential difference between the plates is increased by 0.1%. Hence, the correct answer is option 'D' (0.2%).

Two charged conducting spheres of radii a and b are connected to each other by a wire. The ratio of electric fields at the surfaces of two spheres is
  • a)
    a/b
  • b)
    b/a
  • c)
    a2/b2
  • d)
    b2/a2
Correct answer is option 'B'. Can you explain this answer?

Mira Joshi answered
Let q1 and q2 be the charges and C1 and C2 be the capacitance of two spheres
The charge flows from the sphere at higher potential to the other at lower potential, till their potentials becomes equal.
After sharing, the charges on two spheres would be
q1/q2 = C1V/C2V…(i)
Also C1C2 = ab…(ii)
From (i) q1/q2 = a/b
Ratio of surface charge on the two spheres
σ1/σ2 = q1/4πa2⋅4πb2/q2 = q1/q2⋅b2/a2 = b/a(using(ii)
∴ The ratio of electric fields at the surfaces of two spheres E1/E2 = σ1/σ2 = b/a

If two spheres of different radii have equal charge, then the potential will be
  • a)
    dependent on nature of material of the spheres.
  • b)
    more on smaller sphere
  • c)
    more on bigger sphere
  • d)
    equal on both spheres
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.

The capacitor preferred for high-frequency circuit is 
  • a)
    Air capacitor 
  • b)
    Mica capacitor
  • c)
    Electrolytic capacitor 
  • d)
    None of the above
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
Mica capacitor. Mica capacitors have low resistive and inductive components associated with it. Hence, they have high Q factor and because of high Q factor their characteristics are mostly frequency independent, which allows this capacitor to work at high frequency.

Chapter doubts & questions for Electrostatic Potential & Capacitance - Physics for JAMB 2025 is part of JAMB exam preparation. The chapters have been prepared according to the JAMB exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JAMB 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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