Question

The maximum electric field can be given by

• A. $E_{max}=\frac{Q}{4\sqrt{3}\pi {ϵ}_0{R}^2}$
• B. $E_{max}=\frac{Q}{3\sqrt{3}\pi {ϵ}_0{R}^2}$
• C. $E_{max}=\frac{Q}{2\sqrt{3}\pi {ϵ}_0{R}^2}$
• D. $E_{max}=\frac{Q}{6\sqrt{3}\pi {ϵ}_0{R}^2}$

Answer 1

The correct option is D $E_{max}=\frac{Q}{6\sqrt{3}\pi {ϵ}_0{R}^2}$

The field at point P is given by

$E=\frac{kQx}{(R^2+x^2{)}^{3/2}}$

$\frac{dE}{dx}=kQ\frac{(R^2+x^2{)}^{3/2}-\frac{3}{2}x(R^2+x^2{)}^{1/2}2x}{(R^2+x^2{)}^3}$

$⟹\frac{dE}{dx}=kQ\frac{\sqrt{R^2+x^2}(R^2+x^2-3x^2)}{(R^2+x^2{)}^3}$

For, E to be maximum, $\frac{dE}{dx}=0$

$⟹R^2-2x^2=0$

$⟹x=\pm \frac{R}{\sqrt{2}}$

Field at that point, $E=\frac{kQ\frac{R}{\sqrt{2}}}{(R^2+R^2/2{)}^{3/2}}$

$⟹E=\frac{2kQ}{3\sqrt{3}{R}^2}$

$⟹E=\frac{Q}{6\sqrt{3}\pi {ϵ}_o{R}^2}$

Answer-(D)