Question

Force of attraction between the plates of a parallel plate capacitor (filled with dielectric of constant $k$) is

• A. $\frac{q^2}{2{\epsilon }_{0}Ak}$
• B. $\frac{q^2}{{\epsilon }_{0}Ak}$
• C. $\frac{q}{2{\epsilon }_{0}Ak}$
• D. $\frac{q^2}{2{\epsilon }_0{A}^{2}k}$

Answer 1

The correct option is A $\frac{q^2}{2{\epsilon }_{0}Ak}$

As we know that the electric force between the plates of capacitor is given by $F=qE$.

The electric field due to one plate is given by (if dielectric medium is present), $E=\frac{\sigma }{2{\epsilon }_{0}k}$

Therefore, force between the plates is
$F=q\left(\frac{\sigma }{2{\epsilon }_{0}k}\right)$

As we know that $\sigma =\frac{q}{A}$

Therefore, force $F=q\left(\frac{q}{A2{\epsilon }_{0}k}\right)=\frac{q^2}{2A{\epsilon }_{0}k}$