A parallel plate capacitor with plate area A and separation d is charged to potential V. If the separation is doubled with the charge constant, the new potential is:
Correct answer: C. 2V
- A. V/2
- B. V
- C. 2V
- D. 4V
Explanation
Capacitance is C = ε₀A/d. With constant charge Q, potential is V = Q/C. If d doubles, C' = C/2, so V' = Q/(C/2) = 2Q/C = 2V.
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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