In a space having electric field vec{E}=A(x h | vedclass

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(C) The relation between electric field $\vec{E}$ and potential $V$ is given by $\vec{E} = -
abla V$,which implies $dV = -\vec{E} \cdot d\vec{r}$.Giv

Vedclass · Accepted Answer

$2500$

Vedclass · Answer

(C) The relation between electric field $\vec{E}$ and potential $V$ is given by $\vec{E} = -
abla V$,which implies $dV = -\vec{E} \cdot d\vec{r}$.Given $\vec{E} = A(x \hat{i} + y \hat{j})$,we have $dV = -A(x dx + y dy)$.Integrating both sides,we get $V = -A \int x dx - A \int y dy = -A \frac{x^2}{2} - A \frac{y^2}{2} + C$,where $C$ is the integration constant.Substituting $A = 10 \ Vm^{-2}$,we get $V = -5(x^2 + y^2) + C$.Given that the potential at $(10 \ m, 20 \ m)$ is zero,we have $0 = -5(10^2 + 20^2) + C$.$0 = -5(100 + 400) + C \Rightarrow 0 = -5(500) + C \Rightarrow C = 2500 \ V$.Thus,the potential function is $V(x, y) = -5(x^2 + y^2) + 2500$.At the origin $(0, 0)$,the potential is $V(0, 0) = -5(0^2 + 0^2) + 2500 = 2500 \ V$.

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

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