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. $\frac{q^{2}}{2 \varepsilon_{0} A K}$
• B. $\frac{q^{2}}{\varepsilon_{0} A K}$
• C. $\frac{q}{2 \varepsilon_{0} A}$
• D. $\frac{q^{2}}{2 \varepsilon_{0} A^{2} K}$

Answer 1

Answer: (A)

Force on one plate due to another is
$F=q E=q \times \frac{\sigma}{2 \varepsilon_{0} K}$
$=q\left(\frac{q}{2 A K \varepsilon_{0}}\right)=\frac{q^{2}}{2 A K \varepsilon_{0}}$
(where $\frac{\sigma}{2 \varepsilon_{0} K}$ is the electric field produced by one plate at the location of other).