Question:medium

The electric field between the plates of a parallel plate capacitor is 'E'. If the charge on the plates is Q then the force on each plate is

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The field of one plate is E/2, and it acts on the charge of the other plate.
Updated On: Oct 1, 2026
  • \(\text{QE}^2\)
  • \(\text{QE}\)
  • \(\frac{\text{QE}}{2}\)
  • \(\frac{\text{QE}^2}{2}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Energy method:
Energy stored: $U = \tfrac12QV = \tfrac12QEd$.

Step 2: Force from energy:
At constant $Q$, force $= \dfrac{dU}{dd} = \dfrac12QE$.

Step 3: Result:
$F = \dfrac{QE}{2}$, option (C).

Final Answer:
The force on each plate is QE / 2. \[ \boxed{\text{(C) }\dfrac{QE}{2}} \]
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