Free Capacitors MCQs with Answers

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.