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Q. Statement I: The force with which one plate of a parallel plate capacitor is attracted towards the other plate is equal to square of surface density per $\varepsilon$ per unit area.
Statement II: The electric field due to one charged plate of the capacitor at the location of the other is equal to surface density per $\varepsilon$.

Electrostatic Potential and Capacitance

Solution:

The electric field due to one charged plate at the location of the other is $E=\sigma / 2 \varepsilon_{0}$ and the force per unit area is $F=\sigma E=\sigma^{2} / 2 \varepsilon_{0}$.