Question

Electric field at a distance $x$ from origin is given as $E=\frac{100}{x^2}\text{N/C}$, then the potential difference between the points situated at $x=10\text{m}$ and $x=20\text{m}$ is

• A. $5\text{V}$
• B. $10\text{V}$
• C. $15\text{V}$
• D. $4\text{V}$

Answer 1

The correct option is A $5\text{V}$

We know that, $E=-\frac{dV}{dx}$
$\Rightarrow {\int }_{V_1}^{V_2}dV=-{\int }_{10}^{20}Edx=-{\int }_{10}^{20}\frac{100}{x^2}dx$
$\Rightarrow V_2-V_1=+100(\frac{1}{20}-\frac{1}{10})$
$\Rightarrow V_2-V_1=+100\times \frac{-1}{20}=-5\text{V}$
$\Rightarrow V_1-V_2=5\text{V}$