In a parallel-plate capacitor with plate area $A$ and Charge $Q$, the force on one plate because of the charge on the other is equal to:
- A$\frac{Q^2}{\varepsilon_0A^2}$
- B$\frac{Q^2}{2\varepsilon_0A^2}$
- C$\frac{Q^2}{\varepsilon_0A}$
- D$\frac{Q^2}{2\varepsilon_0A}$
✓ Correct answer: D
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