Free Electrostatics MCQs with Answers

11. Capacitance is defined as

• A. the charge stored per unit potential difference across the capacitor
• B. the energy stored per unit charge
• C. the charge multiplied by the voltage
• D. the potential difference per unit charge

Explanation: C equals Q over V, measured in farads, and it depends only on the geometry of the plates and the dielectric between them, not on how much charge happens to be stored. A farad is a very large unit, so practical capacitors are rated in microfarads or picofarads. Doubling the voltage doubles the charge stored and leaves the capacitance unchanged.

12. The capacitance of a parallel plate capacitor is increased by

• A. increasing the plate separation
• B. decreasing the plate area
• C. increasing the plate area and inserting a dielectric
• D. increasing the applied voltage

Explanation: Capacitance is proportional to the area and to the relative permittivity of the dielectric, and inversely proportional to the separation, so larger closer plates with a dielectric between them store more charge per volt. Applied voltage does not appear in the expression at all, which is the trap in the last option. A dielectric also allows a higher working voltage before breakdown.

13. Three capacitors are connected in parallel. The total capacitance is

• A. the sum of the individual capacitances
• B. less than the smallest of them
• C. the reciprocal of the sum of the reciprocals
• D. always equal to one of them

Explanation: Parallel capacitors share the same voltage and their plate areas effectively add, so the capacitances add directly. In series the reciprocals add, giving a total smaller than the smallest capacitor. This is exactly the opposite of how resistors behave, which is why the two are so often confused.

14. The energy stored in a charged capacitor is given by

• A. QV
• B. half CV squared
• C. CV
• D. half QC

Explanation: The factor of one half appears because the voltage rises from zero to V as the capacitor charges, so the average voltage during the process is half the final value. The expression can equally be written as half QV or as Q squared over 2C. This stored energy is what a camera flash releases in a few milliseconds.

15. A dielectric inserted between the plates of a charged, isolated capacitor causes the potential difference to

• A. increase
• B. decrease, because the capacitance rises while the charge is fixed
• C. stay the same
• D. become zero

Explanation: With the capacitor disconnected the charge cannot change, and since V equals Q over C, raising the capacitance must lower the voltage. The dielectric molecules polarise and set up a field opposing the original one, which is the physical reason for the drop. If instead the capacitor stayed connected to a battery, the voltage would be fixed and the charge would rise.

16. Two identical charges are placed a fixed distance apart. The point where the resultant field is zero lies

• A. midway between them
• B. at one of the charges
• C. outside the pair, beyond the smaller charge
• D. nowhere, since the field is never zero

Explanation: For like charges of equal magnitude the two fields are equal and opposite at the midpoint, so they cancel there. If the charges were unlike, no point between them could give cancellation because both fields would point the same way, and the null point would lie outside the pair on the side of the smaller charge. Working out which case applies is the first step in any such question.

17. Charge on a charged conductor resides

• A. uniformly throughout its volume
• B. on its outer surface, concentrating most where the surface is sharply curved
• C. at its centre
• D. only where it was first placed

Explanation: Mutual repulsion drives the charges as far apart as possible, so they collect on the outside, and the surface charge density is highest at points and edges. This concentration produces the very strong field that ionises the surrounding air at a sharp point, which is the principle of the lightning conductor. Inside the metal the field remains zero.

18. The relationship between uniform electric field intensity E and potential difference V across a separation d is

• A. E equals V times d
• B. E equals V divided by d
• C. E equals d divided by V
• D. E equals V plus d

Explanation: For a uniform field, such as that between parallel plates, the intensity is the potential gradient, so a 100 V supply across plates 2 mm apart gives a field of 50,000 V per metre. This is why the unit volt per metre is equivalent to newton per coulomb. In a non uniform field the same relationship holds only as a local rate of change.

19. Two protons are brought closer together. The electrostatic potential energy of the pair

• A. increases, because work must be done against their repulsion
• B. decreases
• C. stays constant
• D. becomes negative

Explanation: Like charges repel, so pushing them together requires work and that work is stored as potential energy, which is why the system flies apart if released. For unlike charges the energy is negative and becomes more negative as they approach, since the force does the work instead. The sign convention takes zero energy to be infinite separation.

20. An electric dipole placed in a uniform electric field experiences

• A. a net force but no torque
• B. a torque but no net force
• C. neither force nor torque
• D. both a large net force and a torque

Explanation: The equal and opposite charges feel equal and opposite forces in a uniform field, so the forces cancel, but because they act along different lines they form a couple that turns the dipole into alignment with the field. In a non uniform field the two forces differ in size and a net force does appear. This turning effect is how polar molecules align in a dielectric.