A parallel plate capacitor has an electric field of \(10^5\) V/m between the plates. If the charge on the capacitor plate is 1 \(\mu\)C, the force on each capacitor plate is
Show Hint
- 0.5 N
- 0.05 N
- 0.005 N
- None of these
The Correct Option is B
Solution and Explanation
Force on plate \(F = qE/2\).
Step 2: Detailed Explanation:
\(F = (1 \times 10^{-6} \times 10^5)/2 = 10^{-1}/2 = 0.05\ \mathrm{N}\)
Step 3: Final Answer:
Force is 0.05 N.
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