Question

Two charges of $50\mu C$ and $100\mu C$ are separated by a distance of $0.6m$. The intensity of electric field at a point midway between them is :

• A. $50\times {10}^{6}V/m$
• B. $5\times {10}^{6}V/m$
• C. $10\times {10}^{6}V/m$
• D. $10\times {10}^{-6}V/m$

Answer 1

Answer: (B)

$5\times {10}^{6}V/m$

Electric field are vector quantities in this case they are acting in opposite direction.
E due to A$=\frac{kq}{r2}=\frac{(9\times {10}^9)\left(50\mu C\right)}{(0\cdot 3{)}^2}\left(\hat{x}\right)$
E due to B$=\frac{(9\times {10}^9)\left(100\mu C\right)}{(0\cdot 3{)}^2}(-\hat{x})$
$\therefore NetE=\frac{9\times {10}^9}{(0\cdot 3{)}^2}(50-100)\times {10}^{-6}\hat{x}$
$E=5\times {10}^6(-\hat{x})V/m$