Potential V of an electrostatic field vec E = aleft( {yhat{i} + xhat{j}} right) , where a

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2. Electric Potential and Capacitance

easy

Find the potential $V$ of an electrostatic field $\vec E = a\left( {y\hat i + x\hat j} \right)$, where $a$ is a constant.

$axy + C$

$-axy + C$

$axy$

$-axy$

Solution

$\mathrm{E}=a(y i+x j)$

$\mathrm{V}_{2}-\mathrm{V}_{1}=-\int(\mathrm{ayd} \mathrm{x}+\mathrm{axd} \mathrm{y})$

Take $\mathrm{V}_{1}=\mathrm{C}$ and $\mathrm{V}_{2}=\mathrm{V} .$ Then

${\mathrm{V}=-\mathrm{a} \int(\mathrm{ydx}+\mathrm{xdy})+\mathrm{C}} $

${=-\mathrm{a} \int \mathrm{d}(\mathrm{yx})+\mathrm{C}=-\mathrm{axy}+\mathrm{C}}$

$\mathrm{O} \mathrm{R}$

Verify with $ \overrightarrow{\mathrm{E}}=-\nabla \mathrm{V}=-\left(\hat{i} \frac{\partial}{\partial \mathrm{x}}+\hat{j} \frac{\partial}{\partial \mathrm{y}}\right)(-\mathrm{axy}+\mathrm{C})$

$=a y \hat{i}+a x \hat{j}+0=a(y \hat{i}+x \hat{j})$

Standard 12

Physics

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medium

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hard

(JEE MAIN-2023)

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medium

(KVPY-2009)

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medium

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medium

Find the potential $V$ of an electrostatic field $\vec E = a\left( {y\hat i + x\hat j} \right)$, where $a$ is a constant.

Solution

Similar Questions

In a uniform electric field, the potential is $10$ $V $ at the origin of coordinates, and $8$ $V$ at each of the points $(1, 0, 0), (0, 1, 0) $ and $(0, 0, 1)$. The potential at the point $(1, 1, 1)$ will be….$V$

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