Question

Two charges $10\mu C$ and $-10\mu C$ are placed at points $A$ and $B$ separated by a distance of $10cm$ .Find the electric field at a point $P$ on the perpendicular bisector of $AB$ at a distance of $12cm$ from its middle point

Answer 1

The net electric field is given as,

$E=2\times \left[\frac{1}{4\pi {\epsilon }_0}\times \frac{Q_1{d}_1}{{\left({\left(d_1\right)}^2+{\left(d_2\right)}^2\right)}^{\frac{3}{2}}}\right]$

$=2\times \left[9\times {10}^9\times \frac{10\times {10}^{-6}\times \left(5\times {10}^{-2}\right)}{{\left({\left(5\times {10}^{-2}\right)}^2+{\left(12\times {10}^{-12}\right)}^2\right)}^{\frac{3}{2}}}\right]$

$=4.096\times {10}^{6}V/m$

Thus, the electric field is $4.096\times {10}^{6}V/m$.