Electrostatic Potential and Capacitance Important Questions.

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1. Q1Medium

   If three capacitors of equal capacitance (C) are connected in series, what is the equivalent capacitance?

   • A.3C
   • B.C
   • C.C/3
   • D.C + C + C
   Answer: C/3

   Explanation: When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. For three equal capacitors, this results in C/3.

2. Q2Medium

   What is the potential difference across a capacitor if the charge on its plates is 10 µC and its capacitance is 2 µF?

   • A.2 V
   • B.5 V
   • C.10 V
   • D.20 V
   Answer: 5 V

   Explanation: Using the formula for potential difference (V = Q/C), where Q is the charge and C is the capacitance, we get V = 10 µC / 2 µF = 5 V.

3. Q3Medium

   Which of the following statements about equipotential surfaces is true?

   • A.Equipotential surfaces are parallel to electric field lines.
   • B.Equipotential surfaces are only found inside conductors.
   • C.Equipotential surfaces are perpendicular to electric field lines.
   • D.Equipotential surfaces can intersect each other.
   Answer: Equipotential surfaces are perpendicular to electric field lines.

   Explanation: Equipotential surfaces are defined as surfaces where the electric potential is constant. They are always perpendicular to the electric field lines.

4. Q4Medium

   What is the electric field inside a conductor in electrostatic equilibrium?

   • A.Non-zero and constant
   • B.Equal to the external field
   • C.Zero
   • D.Infinite
   Answer: Zero

   Explanation: In electrostatic equilibrium, the electric field inside a conductor is zero because any net field would cause the free charges to move until equilibrium is reached.

5. Q5Medium

   What happens to the capacitance of a parallel plate capacitor if the distance between the plates is doubled?

   • A.The capacitance doubles.
   • B.The capacitance remains the same.
   • C.The capacitance is halved.
   • D.The capacitance becomes infinite.
   Answer: The capacitance is halved.

   Explanation: The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. Doubling the distance halves the capacitance.

6. Q6Medium

   What is the potential energy stored in a capacitor with capacitance 5 µF and potential difference 10 V?

   • A.100 µJ
   • B.250 µJ
   • C.500 µJ
   • D.1000 µJ
   Answer: 250 µJ

   Explanation: Using the formula for energy stored in a capacitor, U = 0.5 * C * V^2, we get U = 0.5 * 5 µF * (10 V)^2 = 250 µJ.

7. Q7Medium

   Which principle is used to determine the total electric potential due to multiple charge distributions?

   • A.Newton’s Law of Gravitation
   • B.Superposition Principle
   • C.Ohm’s Law
   • D.Faraday’s Law
   Answer: Superposition Principle

   Explanation: The superposition principle states that the total potential at any point due to multiple charges is the algebraic sum of the potentials due to each individual charge.

8. Q8Medium

   What is the dielectric constant of a material if the capacitance increases by a factor of 6 when it is inserted between the plates of a parallel plate capacitor?

   • A.2
   • B.4
   • C.6
   • D.8
   Answer: 6

   Explanation: The dielectric constant (κ) is defined as the ratio of the capacitance with the dielectric to the capacitance without the dielectric. If the capacitance increases by a factor of 6, the dielectric constant is 6.

9. Q9Medium

   What is the unit of electric potential?

   • A.Joule
   • B.Coulomb
   • C.Volt
   • D.Ampere
   Answer: Volt

   Explanation: The SI unit of electric potential is the volt (V), which is equivalent to one joule per coulomb.

10. Q10Medium

    What is the potential energy of an electron (charge -e) in an electric potential of 5 V?

    • A.5 eV
    • B.-1.6 x 10^-19 J
    • C.1.6 x 10^-19 J
    • D.8 eV
    Answer: -1.6 x 10^-19 J

    Explanation: The potential energy is given by U = qV, where q is the charge of the electron (-1.6 x 10^-19 C) and V is the potential (5 V). Thus, U = -1.6 x 10^-19 C * 5 V = -8 x 10^-19 J (approximately -1.6 x 10^-19 J in electron volts).

11. Q11Medium

    What is the electric field inside a dielectric material between the plates of a capacitor?

    • A.Increased compared to the field without the dielectric.
    • B.Same as the field without the dielectric.
    • C.Reduced compared to the field without the dielectric.
    • D.Zero.
    Answer: Reduced compared to the field without the dielectric.

    Explanation: The presence of a dielectric reduces the electric field between the plates of a capacitor due to polarization effects, which oppose the applied field.

12. Q12Medium

    What is the work done to move a charge of 2 µC through a potential difference of 10 V?

    • A.5 µJ
    • B.10 µJ
    • C.20 µJ
    • D.50 µJ
    Answer: 20 µJ

    Explanation: The work done (W) is given by W = qV, where q is the charge (2 µC) and V is the potential difference (10 V). Thus, W = 2 µC * 10 V = 20 µJ.

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