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A parallel plate capacitor is charged with a battery and then separated from it. Now if the distance between its two plates is increased, what will be the changes in electric charge, potential difference and capacitance respectively?

Asked in RS Academy GUJCET booklet · Battery connected or removed

• (1) remains constant, decreases, decreases
• (2) increases, decreases, decreases
• (3) remains constant, decreases, increases
• (4) remains constant, increases, decreases

Step-by-step solution

Why the other options are wrong

• (1) Charge and capacitance are right, but at fixed charge V = Q/C must rise when C falls; it cannot decrease.
• (2) The capacitor is disconnected, so no charge can flow on or off its plates: Q stays constant.
• (3) Has the capacitance the wrong way round: C = (ε₀A)/d falls as d grows. (With C taken as rising, V = Q/C then comes out falling — also wrong.)

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