A parallel-plate capacitor has a plate area of \(0.3\text{ m}^2\) and a plate separation of \(0.1\text{ mm}\). If the charge on each plate has a magnitude of \(5 \times 10^{-6}\text{ C}\) then the force exerted by one plate on the other has a magnitude of about :
\(1 \times 10^4\text{ N}\)
\(9 \times 10^5\text{ N}\)
Solution:
The force of attraction between the plates of a parallel plate capacitor is given by \(F = \frac{Q^2}{2 \epsilon_0 A}\). Substituting the values: \(F = \frac{(5 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12} \times 0.3} \approx 4.7\text{ N}\), which is about \(5\text{ N}\).
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