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In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$ the potential at a point $(10 \mathrm{~m}, 20 \mathrm{~m})$ is zero, then the potential at the origin is $\left[\mathrm{A}=10 \mathrm{Vm}^{-2}\right]$
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The correct answer is:
$2500 \mathrm{~V}$
We have,
$\begin{aligned} & \vec{E}=A(x \hat{i}+y \hat{j}) \\ & V=-\int E d r=-A \int x d x-A \int y d y\end{aligned}$
$\begin{aligned} & =-\mathrm{A} \frac{\mathrm{x}^2}{2}-\mathrm{A} \frac{\mathrm{y}^2}{2}+\mathrm{V}_0 \\ & \mathrm{~V}_0=\text { Integration constant }\end{aligned}$
$=-\frac{A}{2}\left(x^2+y^2\right)+V_0=-5\left(x^2+y^2\right)+V_0$
$\mathrm{V|}_{(10,20)}=0 \Rightarrow-5\left(10^2+20^2\right)+\mathrm{V}_0=0 \Rightarrow \mathrm{V}_0=2500$
So, $\left.\mathrm{V}\right|_{\mathrm{x}=0, \mathrm{y}=0}=-5\left(0^2+0^2\right)+2500=2500 \mathrm{~V}$
$\begin{aligned} & \vec{E}=A(x \hat{i}+y \hat{j}) \\ & V=-\int E d r=-A \int x d x-A \int y d y\end{aligned}$
$\begin{aligned} & =-\mathrm{A} \frac{\mathrm{x}^2}{2}-\mathrm{A} \frac{\mathrm{y}^2}{2}+\mathrm{V}_0 \\ & \mathrm{~V}_0=\text { Integration constant }\end{aligned}$
$=-\frac{A}{2}\left(x^2+y^2\right)+V_0=-5\left(x^2+y^2\right)+V_0$
$\mathrm{V|}_{(10,20)}=0 \Rightarrow-5\left(10^2+20^2\right)+\mathrm{V}_0=0 \Rightarrow \mathrm{V}_0=2500$
So, $\left.\mathrm{V}\right|_{\mathrm{x}=0, \mathrm{y}=0}=-5\left(0^2+0^2\right)+2500=2500 \mathrm{~V}$
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