As the plates of a charged variable capacitor are moved closer together, the potential difference between them
A
increased
B
decrease
C
remains the same
D
is doubled
Verified Correct Answer: B.
Physical Principles & Numerical Solution
As the plates of a charged variable capacitor are moved closer together, the potential difference between them decreases.
This is because the capacitance increases as the distance between the plates decreases. For a constant charge, the potential difference (V = \(\frac{\text{Q}}{\text{C}}\)) will decrease when capacitance (C) increases.
Source: Myschool Past Questions Archive (Record #33934)