Question
Establish the relation between electric field and potential gradient.
✓
Answer
Let us consider two closely spaced equipotential surfaces A and B as shown in figure.
Let the potential of A be $V_A = V $ and potential of B be $V_B = V - dV$
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 $V_A - V_B,$
$\therefore\text{W}_{\text{BA}}=\text{V}_{\text{A}}-\text{V}_{\text{B}}=\text{V}-(\text{V}-\text{dV})=\text{dV}$
$\therefore-\vec{\text{E}}=\text{dV}$ Or, $\vec{\text{E}}=-\frac{\text{dV}}{\text{dr}}$
= negative of potential gradlant.
Let the potential of A be $V_A = V $ and potential of B be $V_B = V - dV$
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 $V_A - V_B,$
$\therefore\text{W}_{\text{BA}}=\text{V}_{\text{A}}-\text{V}_{\text{B}}=\text{V}-(\text{V}-\text{dV})=\text{dV}$
$\therefore-\vec{\text{E}}=\text{dV}$ Or, $\vec{\text{E}}=-\frac{\text{dV}}{\text{dr}}$
= negative of potential gradlant.
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