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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance5 Marks
Question
Establish the relation between electric field and potential gradient.
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Answer

Let us consider two closely spaced equipotential surfaces A and B as shown in figure.

Let the potential of A be VA = V and potential of B be VB = V - dV where dV is decrease in potential in the direction of electric field $\vec{\text{E}}$ normal to A and B.

Let dr be the perpendicular distance between the two equipotential surfaces. When a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field, the work done in this process is:

$\text{W}_{\text{BA}}=-\vec{\text{E}}(\text{dr})$

This work done equals the potentail difference VA - VB,

$\therefore\text{W}_{\text{BA}}=\text{V}_{\text{A}}-\text{V}_{\text{B}}=\text{V}-(\text{V}-\text{dV})=\text{dV}$

$\therefore-\vec{\text{E}}=\text{dV}$

Or, $\vec{\text{E}}=-\frac{\text{dV}}{\text{dr}}$ = negative of potential gradlant.

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