Study for your Class 12 Physics exams with these MCQs on Electrostatic Potential and Capacitance. Based on the latest CBSE syllabus and NCERT, these questions cover important topics to help you get ready for your exams.
Electrostatic Potential and Capacitance MCQ
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1. The electrostatic potential \( V \) at a distance \( r \) from a point charge \( Q \) in a vacuum is given by:
A. \( \frac{Q}{4 \pi \varepsilon_0 r^2} \)
B. \( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \)
C. \( \frac{4 \pi \varepsilon_0 Q}{r} \)
D. \( \frac{Q r}{4 \pi \varepsilon_0} \)
The correct answer is B. \( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \).
Understanding Electrostatic Potential:
The electrostatic potential \( V \) due to a point charge \( Q \) at a distance \( r \) is defined as the work done in bringing a unit positive charge from infinity to that point in the electric field of \( Q \).
Formula for Electrostatic Potential:
The formula for the electrostatic potential at distance \( r \) from a charge \( Q \) is: \[ V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \] where: - \( \varepsilon_0 \) is the permittivity of free space, with a value of \( 8.854 \times 10^{-12} \, \text{F/m} \). - \( Q \) is the point charge. - \( r \) is the distance from the charge \( Q \) to the point where the potential is being calculated.
Conclusion:
This formula shows that the electrostatic potential decreases with distance from the charge and is directly proportional to the magnitude of the charge \( Q \).
Understanding Electrostatic Potential:
The electrostatic potential \( V \) due to a point charge \( Q \) at a distance \( r \) is defined as the work done in bringing a unit positive charge from infinity to that point in the electric field of \( Q \).
Formula for Electrostatic Potential:
The formula for the electrostatic potential at distance \( r \) from a charge \( Q \) is: \[ V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \] where: - \( \varepsilon_0 \) is the permittivity of free space, with a value of \( 8.854 \times 10^{-12} \, \text{F/m} \). - \( Q \) is the point charge. - \( r \) is the distance from the charge \( Q \) to the point where the potential is being calculated.
Conclusion:
This formula shows that the electrostatic potential decreases with distance from the charge and is directly proportional to the magnitude of the charge \( Q \).
2. The electrostatic potential energy \( U \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
A. \( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \)
B. \( \frac{q_1 q_2}{4 \pi \varepsilon_0 r^2} \)
C. \( \frac{q_1 q_2 r}{4 \pi \varepsilon_0} \)
D. \( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r} \)
The correct answer is A. \( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \).
Understanding Electrostatic Potential Energy:
The electrostatic potential energy between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is the work required to assemble these charges at that separation.
Formula for Electrostatic Potential Energy:
The formula for potential energy \( U \) is: \[ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, - \( r \) is the distance separating the charges.
Conclusion:
This equation shows that the potential energy depends on both charges and the distance between them, decreasing as \( r \) increases.
Understanding Electrostatic Potential Energy:
The electrostatic potential energy between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is the work required to assemble these charges at that separation.
Formula for Electrostatic Potential Energy:
The formula for potential energy \( U \) is: \[ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, - \( r \) is the distance separating the charges.
Conclusion:
This equation shows that the potential energy depends on both charges and the distance between them, decreasing as \( r \) increases.
3. Which of the following correctly describes an equipotential surface for a point charge?
A. Concentric cylinders around the charge
B. Concentric spheres centered at the charge
C. Planes perpendicular to the charge
D. Randomly shaped surfaces
The correct answer is B. Concentric spheres centered at the charge.
Understanding Equipotential Surfaces:
An equipotential surface is a surface where the potential at every point is the same. For a point charge, the equipotential surfaces are spherical because the potential only depends on the distance from the charge, not the direction.
Explanation:
Since the electric field radiates outward uniformly in all directions from a point charge, the equipotential surfaces form concentric spheres around the charge.
Conclusion:
Hence, the correct shape for an equipotential surface around a point charge is a sphere centered at the charge.
Understanding Equipotential Surfaces:
An equipotential surface is a surface where the potential at every point is the same. For a point charge, the equipotential surfaces are spherical because the potential only depends on the distance from the charge, not the direction.
Explanation:
Since the electric field radiates outward uniformly in all directions from a point charge, the equipotential surfaces form concentric spheres around the charge.
Conclusion:
Hence, the correct shape for an equipotential surface around a point charge is a sphere centered at the charge.
4. The capacitance \( C \) of a parallel plate capacitor with vacuum between the plates is given by:
A. \( C = \frac{\varepsilon_0 A}{d} \)
B. \( C = \frac{\varepsilon_0 d}{A} \)
C. \( C = \varepsilon_0 A d \)
D. \( C = \frac{A}{\varepsilon_0 d} \)
The correct answer is A. \( C = \frac{\varepsilon_0 A}{d} \).
Understanding Capacitance of a Parallel Plate Capacitor:
Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. For a parallel plate capacitor, the capacitance depends on the area of the plates \( A \) and the separation \( d \) between them.
Formula for Capacitance:
The formula for capacitance \( C \) with vacuum between the plates is: \[ C = \frac{\varepsilon_0 A}{d} \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( A \) is the area of each plate, - \( d \) is the separation between the plates.
Conclusion:
Capacitance increases with plate area and decreases with plate separation.
Understanding Capacitance of a Parallel Plate Capacitor:
Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. For a parallel plate capacitor, the capacitance depends on the area of the plates \( A \) and the separation \( d \) between them.
Formula for Capacitance:
The formula for capacitance \( C \) with vacuum between the plates is: \[ C = \frac{\varepsilon_0 A}{d} \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( A \) is the area of each plate, - \( d \) is the separation between the plates.
Conclusion:
Capacitance increases with plate area and decreases with plate separation.
5. For a series combination of capacitors with capacitances \( C_1, C_2, \) and \( C_3 \), the total capacitance \( C_{\text{total}} \) is given by:
A. \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \)
B. \( C_{\text{total}} = C_1 + C_2 + C_3 \)
C. \( C_{\text{total}} = \frac{C_1 C_2 C_3}{C_1 + C_2 + C_3} \)
D. \( C_{\text{total}} = \frac{C_1 + C_2 + C_3}{3} \)
The correct answer is A. \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \).
Understanding Series Capacitance:
When capacitors are connected in series, the total capacitance is less than any individual capacitance in the circuit because the effective separation between charges increases.
Formula for Series Capacitance:
For \( n \) capacitors in series, the total capacitance \( C_{\text{total}} \) is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots \] where \( C_1, C_2, C_3, \dots \) are the individual capacitances.
Conclusion:
In a series combination, the total capacitance decreases as more capacitors are added.
Understanding Series Capacitance:
When capacitors are connected in series, the total capacitance is less than any individual capacitance in the circuit because the effective separation between charges increases.
Formula for Series Capacitance:
For \( n \) capacitors in series, the total capacitance \( C_{\text{total}} \) is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots \] where \( C_1, C_2, C_3, \dots \) are the individual capacitances.
Conclusion:
In a series combination, the total capacitance decreases as more capacitors are added.
6. The energy \( U \) stored in a capacitor of capacitance \( C \) charged to a voltage \( V \) is given by:
A. \( U = \frac{1}{2} C V^2 \)
B. \( U = CV \)
C. \( U = C V^2 \)
D. \( U = \frac{C}{2} V \)
The correct answer is A. \( U = \frac{1}{2} C V^2 \).
Understanding Energy in a Capacitor:
The energy stored in a capacitor arises from the work done to move charge onto the plates, creating an electric field.
Formula for Stored Energy:
The energy \( U \) stored in a capacitor with capacitance \( C \) and voltage \( V \) across its plates is: \[ U = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the potential difference across the plates.
Conclusion:
This expression shows that the energy stored depends on both the capacitance and the square of the voltage applied to the capacitor.
Understanding Energy in a Capacitor:
The energy stored in a capacitor arises from the work done to move charge onto the plates, creating an electric field.
Formula for Stored Energy:
The energy \( U \) stored in a capacitor with capacitance \( C \) and voltage \( V \) across its plates is: \[ U = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance, - \( V \) is the potential difference across the plates.
Conclusion:
This expression shows that the energy stored depends on both the capacitance and the square of the voltage applied to the capacitor.
7. Which of the following statements is true for an equipotential surface?
A. The potential varies along an equipotential surface.
B. The electric field is always parallel to the equipotential surface.
C. Equipotential surfaces can cross each other.
D. The electric field is always perpendicular to the equipotential surface.
The correct answer is D. The electric field is always perpendicular to the equipotential surface..
Understanding Equipotential Surfaces:
An equipotential surface has the same potential at every point, so no work is needed to move a charge along it.
Relation to Electric Field:
The electric field direction is always perpendicular to the equipotential surface. This is because work would be required if it were not, contradicting the nature of equipotential surfaces.
Conclusion:
Electric field lines are perpendicular to equipotential surfaces, as this minimizes the work done on charges moving along these surfaces.
Understanding Equipotential Surfaces:
An equipotential surface has the same potential at every point, so no work is needed to move a charge along it.
Relation to Electric Field:
The electric field direction is always perpendicular to the equipotential surface. This is because work would be required if it were not, contradicting the nature of equipotential surfaces.
Conclusion:
Electric field lines are perpendicular to equipotential surfaces, as this minimizes the work done on charges moving along these surfaces.
8. When a dielectric material is placed between the plates of a capacitor, its capacitance:
A. Increases by a factor equal to the dielectric constant \( K \)
B. Decreases by a factor equal to the dielectric constant \( K \)
C. Remains the same
D. Becomes zero
The correct answer is A. Increases by a factor equal to the dielectric constant \( K \).
Effect of Dielectric Material:
A dielectric material reduces the electric field within the capacitor, allowing more charge to be stored at the same voltage.
Formula for Capacitance with Dielectric:
If \( C_0 \) is the capacitance without the dielectric, then with a dielectric of constant \( K \): \[ C = K C_0 \] where \( K \) is the dielectric constant of the material.
Conclusion:
The presence of a dielectric increases capacitance, enhancing the charge storage capability of the capacitor.
Effect of Dielectric Material:
A dielectric material reduces the electric field within the capacitor, allowing more charge to be stored at the same voltage.
Formula for Capacitance with Dielectric:
If \( C_0 \) is the capacitance without the dielectric, then with a dielectric of constant \( K \): \[ C = K C_0 \] where \( K \) is the dielectric constant of the material.
Conclusion:
The presence of a dielectric increases capacitance, enhancing the charge storage capability of the capacitor.
9. The capacitance \( C \) of a spherical conductor with radius \( R \) in vacuum is given by:
A. \( C = \frac{\varepsilon_0}{4 \pi R} \)
B. \( C = \frac{4 \pi \varepsilon_0}{R} \)
C. \( C = 4 \pi \varepsilon_0 R \)
D. \( C = \varepsilon_0 R^2 \)
The correct answer is C. \( C = 4 \pi \varepsilon_0 R \).
Capacitance of a Spherical Conductor:
For a single isolated spherical conductor in vacuum, capacitance depends solely on its radius and the permittivity of free space.
Formula:
The capacitance \( C \) of a spherical conductor with radius \( R \) is: \[ C = 4 \pi \varepsilon_0 R \] where \( \varepsilon_0 \) is the permittivity of free space and \( R \) is the radius of the sphere.
Conclusion:
The capacitance of a spherical conductor increases linearly with its radius.
Capacitance of a Spherical Conductor:
For a single isolated spherical conductor in vacuum, capacitance depends solely on its radius and the permittivity of free space.
Formula:
The capacitance \( C \) of a spherical conductor with radius \( R \) is: \[ C = 4 \pi \varepsilon_0 R \] where \( \varepsilon_0 \) is the permittivity of free space and \( R \) is the radius of the sphere.
Conclusion:
The capacitance of a spherical conductor increases linearly with its radius.
10. Electrostatic shielding in a conductor implies that:
A. The electric field inside a conductor is zero.
B. The electric field inside a conductor is maximum.
C. Charges reside inside the conductor.
D. Potential is zero at all points outside the conductor.
The correct answer is A. The electric field inside a conductor is zero..
Understanding Electrostatic Shielding:
Electrostatic shielding occurs because any electric field within a conductor induces charges on its surface, which rearrange to cancel the field inside.
Explanation:
In electrostatic equilibrium, the electric field within a conductor is zero, and any net charge resides on the surface. This property allows conductors to shield enclosed regions from external electric fields.
Conclusion:
Electrostatic shielding is a key feature of conductors, making the electric field zero within their interior.
Understanding Electrostatic Shielding:
Electrostatic shielding occurs because any electric field within a conductor induces charges on its surface, which rearrange to cancel the field inside.
Explanation:
In electrostatic equilibrium, the electric field within a conductor is zero, and any net charge resides on the surface. This property allows conductors to shield enclosed regions from external electric fields.
Conclusion:
Electrostatic shielding is a key feature of conductors, making the electric field zero within their interior.
11. The potential energy \( U \) of an electric dipole of dipole moment \( p \) in a uniform electric field \( E \) is given by:
A. \( U = -p \cdot E \cos \theta \)
B. \( U = p \cdot E \sin \theta \)
C. \( U = p \cdot E \)
D. \( U = -p \cdot E \)
The correct answer is A. \( U = -p \cdot E \cos \theta \).
Understanding Potential Energy of a Dipole in a Uniform Electric Field:
The potential energy of a dipole depends on the angle \( \theta \) between the dipole moment \( p \) and the electric field \( E \).
Formula for Potential Energy:
The potential energy \( U \) of a dipole in an electric field is: \[ U = -p \cdot E \cos \theta \] where: - \( p \) is the dipole moment, - \( E \) is the magnitude of the electric field, - \( \theta \) is the angle between \( p \) and \( E \).
Conclusion:
This formula shows that the energy is minimized when the dipole aligns with the field (\( \theta = 0^\circ \)).
Understanding Potential Energy of a Dipole in a Uniform Electric Field:
The potential energy of a dipole depends on the angle \( \theta \) between the dipole moment \( p \) and the electric field \( E \).
Formula for Potential Energy:
The potential energy \( U \) of a dipole in an electric field is: \[ U = -p \cdot E \cos \theta \] where: - \( p \) is the dipole moment, - \( E \) is the magnitude of the electric field, - \( \theta \) is the angle between \( p \) and \( E \).
Conclusion:
This formula shows that the energy is minimized when the dipole aligns with the field (\( \theta = 0^\circ \)).
12. The capacitance of a capacitor with a dielectric material of dielectric constant \( K \) inserted between the plates is given by:
A. \( C = \frac{C_0}{K} \)
B. \( C = K C_0 \)
C. \( C = K^2 C_0 \)
D. \( C = C_0 \)
The correct answer is B. \( C = K C_0 \).
Effect of Dielectric Material on Capacitance:
When a dielectric material is introduced, it increases the capacitance by reducing the effective electric field within the capacitor.
Formula:
The new capacitance \( C \) with dielectric constant \( K \) is: \[ C = K C_0 \] where: - \( C_0 \) is the capacitance without the dielectric.
Conclusion:
A dielectric increases capacitance, allowing more charge to be stored at a given voltage.
Effect of Dielectric Material on Capacitance:
When a dielectric material is introduced, it increases the capacitance by reducing the effective electric field within the capacitor.
Formula:
The new capacitance \( C \) with dielectric constant \( K \) is: \[ C = K C_0 \] where: - \( C_0 \) is the capacitance without the dielectric.
Conclusion:
A dielectric increases capacitance, allowing more charge to be stored at a given voltage.
13. The energy density \( u \) in an electric field \( E \) is given by:
A. \( u = \frac{1}{2} \varepsilon_0 E^2 \)
B. \( u = \frac{1}{2} \varepsilon_0 E \)
C. \( u = \frac{1}{4} \varepsilon_0 E^2 \)
D. \( u = \frac{1}{4} \varepsilon_0 E \)
The correct answer is A. \( u = \frac{1}{2} \varepsilon_0 E^2 \).
Understanding Energy Density in an Electric Field:
Energy density refers to the amount of energy stored per unit volume in an electric field.
Formula:
The energy density \( u \) is: \[ u = \frac{1}{2} \varepsilon_0 E^2 \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( E \) is the electric field.
Conclusion:
The energy density increases with the square of the electric field magnitude.
Understanding Energy Density in an Electric Field:
Energy density refers to the amount of energy stored per unit volume in an electric field.
Formula:
The energy density \( u \) is: \[ u = \frac{1}{2} \varepsilon_0 E^2 \] where: - \( \varepsilon_0 \) is the permittivity of free space, - \( E \) is the electric field.
Conclusion:
The energy density increases with the square of the electric field magnitude.
14. Which of the following correctly describes the behavior of capacitors in a parallel combination?
A. The total capacitance is half the sum of individual capacitances.
B. The total capacitance is the product of individual capacitances.
C. The total capacitance is the reciprocal of the sum of reciprocals.
D. The total capacitance is the sum of individual capacitances.
The correct answer is D. The total capacitance is the sum of individual capacitances..
Capacitors in Parallel:
In a parallel combination, each capacitor experiences the same potential difference. The total capacitance increases because the effective area for charge storage increases.
Formula:
For capacitors \( C_1, C_2, \dots, C_n \) in parallel: \[ C_{\text{total}} = C_1 + C_2 + \dots + C_n \]
Conclusion:
Total capacitance in a parallel arrangement is simply the sum of individual capacitances.
Capacitors in Parallel:
In a parallel combination, each capacitor experiences the same potential difference. The total capacitance increases because the effective area for charge storage increases.
Formula:
For capacitors \( C_1, C_2, \dots, C_n \) in parallel: \[ C_{\text{total}} = C_1 + C_2 + \dots + C_n \]
Conclusion:
Total capacitance in a parallel arrangement is simply the sum of individual capacitances.
15. In electrostatic shielding, if there is no charge inside a conducting cavity, then:
A. The electric field inside the cavity is zero regardless of external fields.
B. The electric field inside the cavity depends on the external field.
C. Charges will flow into the cavity from the conductor.
D. The potential inside the cavity varies with the external field.
The correct answer is A. The electric field inside the cavity is zero regardless of external fields..
Electrostatic Shielding:
In a conductor, any external electric field causes charges to rearrange on its surface, creating an induced field that cancels the field inside the conductor.
Explanation:
Since the cavity is shielded by the conductor, no external field can penetrate inside, resulting in a zero electric field within the cavity if no charge is present.
Conclusion:
Electrostatic shielding ensures that regions inside a conductor remain unaffected by external fields.
Electrostatic Shielding:
In a conductor, any external electric field causes charges to rearrange on its surface, creating an induced field that cancels the field inside the conductor.
Explanation:
Since the cavity is shielded by the conductor, no external field can penetrate inside, resulting in a zero electric field within the cavity if no charge is present.
Conclusion:
Electrostatic shielding ensures that regions inside a conductor remain unaffected by external fields.
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