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 ε 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 ε.

Question

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 ε 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 ε. (A) If both statements are true and statement II is the correct explanation of statement I. (B) If both statements are true but statement II is not the correct explanation of statement I. (C) If statement I is true but statement II is false. (D) If both statements are false.

Tardigrade Admin · Accepted Answer

The electric field due to one charged plate at the location of the other is E=σ / 2 ε0 and the force per unit area is F=σ E=σ2 / 2 ε0.

If both statements are true and statement II is the correct explanation of statement I.

If both statements are true but statement II is not the correct explanation of statement I.

If statement I is true but statement II is false.

If both statements are false.

The electric field due to one charged plate at the location of the other is E=σ/2ε0​ and the force per unit area is F=σE=σ2/2ε0​.

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