Physics - Electrostatics Question with Solution | TestHub

Let the null point be at a distance $r$ to the right of point B.

The electric field due to $+10\text{ q}$ at A ($x=0$) and $-4\text{ q}$ at B ($x=L$) must be equal in magnitude and opposite in direction.

Since the charges have opposite signs, the null point lies outside the region between them. It must be closer to the charge with the smaller magnitude ($-4\text{ q}$). Thus, the null point is to the right of B.

Equating the magnitudes of the electric fields at this point:

$\frac{k(10\text{q})}{(L+r{)}^2}=\frac{k(4\text{q})}{r^2}$

$\frac{5}{(L+r{)}^2}=\frac{2}{r^2}$

Taking the square root of both sides:

$\frac{\sqrt{5}}{L+r}=\frac{\sqrt{2}}{r}$

$\sqrt{5}r=\sqrt{2}L+\sqrt{2}r$

$r(\sqrt{5}-\sqrt{2})=\sqrt{2}L$

$r=\frac{\sqrt{2}L}{\sqrt{5}-\sqrt{2}}$

The point is at a distance $r$ to the right of point B.

Answer: $r=\frac{\sqrt{2}\mathrm{L}}{\sqrt{5}-\sqrt{2}}$ right to pt.B