Question

Electric field $\overrightarrow{E}=20\hat{i}+30\hat{j}N/C$ exist in the space. If the potential at the origin is taken to be zero. Then the potential at $(2m,2m)$ is

• A. $-100V$
• B. $100V$
• C. $200V$
• D. $-200V$

Answer 1

The correct option is A $-100V$

$\text{Step 1: Using the equation}E=-▽V$

$\overrightarrow{E}=\frac{-\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}$ $....\left(1\right)$

$\overrightarrow{E}=20\hat{i}+30\hat{j}N/c$ $....\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, We get:

$\frac{-\partial V}{\partial x}=20$ $....\left(3\right)$

$\frac{-\partial V}{\partial y}=30$ $....\left(4\right)$

$\text{Step 2: Take}\partial x\text{and}\partial y\text{on RHS and do the integration}$

At $x=0,y=0,V=0$ (Given)

From eq. $\left(3\right)$

$-\partial V=20\partial x$

${\int }_0^{V_x}-\partial V={\int }_0^{2}20\partial x$

$\Rightarrow -V_x+0=20(2-0)$

$\Rightarrow V_x=-40$

From eq. $\left(4\right)$

${\int }_0^{V_y}-\partial V={\int }_0^{2}30\partial y$

$\Rightarrow -V_y=30\times 2$

$\Rightarrow V_y=-60$

$\text{Step 3: Potential at (2m , 2m)}$

$V=V_x+V_y$ $=-40-60=-100V$

Hence , Option $\left(A\right)$ is correct.