Force of attraction between the plates of a parallel plate capacitor is (Here, q= charge on plates, A= Area of plates, K= dielectric constant)

Question

Force of attraction between the plates of a parallel plate capacitor is (Here, q= charge on plates, A= Area of plates, K= dielectric constant) (A) (q2/2 ε0 A K) (B) (q2/ε0 A K) (C) (q/2 ε0 A) (D) (q2/2 ε0 A2 K)

Tardigrade Admin · Accepted Answer

Force on one plate due to another is F=q E=q × (σ/2 ε0 K) =q((q/2 A K ε0))=(q2/2 A K ε0) (where (σ/2 ε0 K) is the electric field produced by one plate at the location of other).

Q. Force of attraction between the plates of a parallel plate capacitor is (Here, $q =$ charge on plates, $A =$ Area of plates, $K =$ dielectric constant)

$\frac{q ^{2}}{2 ε _{0} A K}$

$\frac{q ^{2}}{ε _{0} A K}$

$\frac{q}{2 ε _{0} A}$

$\frac{q ^{2}}{2 ε _{0} A ^{2} K}$

Force on one plate due to another is
$F = qE = q \times \frac{σ}{2 ε _{0} K}$
$= q (\frac{q}{2 A K ε _{0}}) = \frac{q ^{2}}{2 A K ε _{0}}$
(where $\frac{σ}{2 ε _{0} K}$ is the electric field produced by one plate at the location of other).

Q. Force of attraction between the plates of a parallel plate capacitor is (Here, q= charge on plates, A= Area of plates, K= dielectric constant)

2ε0​AKq2​

ε0​AKq2​

2ε0​Aq​

2ε0​A2Kq2​