Question

Two charges $+5\mu \text{C}$ and $+10\mu \text{C}$ are placed $20\text{cm}$ apart. The electric field at the mid-point between the two charges will be:

• A. $12.5\times {10}^6\text{N/C}\text{towards}+10\mu \text{C}$
• B. $12.5\times {10}^6\text{N/C}\text{towards}+5\mu \text{C}$
• C. $4.5\times {10}^6\text{N/C}\text{towards}+10\mu \text{C}$
• D. $4.5\times {10}^6\text{N/C}\text{towards}+5\mu \text{C}$

Answer 1

The correct option is D $4.5\times {10}^6\text{N/C}\text{towards}+5\mu \text{C}$

Given: $q_1=10\mu \text{C}$ and $q_2=5\mu \text{C}$

Let the electric field at point $P$ due to $q_1$ is

$E_1$ and due to $q_2$ is $E_2$.

$E_p=E_1-E_2=\frac{q_1}{4\pi {\epsilon }_0(0.1{)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{(0.1{)}^2}$

$E_p=\frac{9\times {10}^9}{(0.1{)}^2}(10-5)\times {10}^{-6}=4.5\times {10}^6\text{N/C}$

As $E_1>E_2$

Electric field is towards $+5\mu \text{C}$.