Name: Dipole in an Electric Field
Uploaded: 2019-07-19
Duration: 11 min 59 s
Description: Dipoles exists everywhere. They apply to both electricity and magnetism in physics. Here in this video, let us explore how an electric dipole interacts with...

Question

Define electric potential and obtain an expression between intensity of electric field due to a point charge and electric potential. Also show that electric field is gradient of electric potential.

Accepted Answer

Electric potential $\to$ It is defined as work done per unit test charge.

$i . e ., [V = \frac{W}{q}]$

Expression for electric field intensity at a point due to point charge $\to$

By Coulomb's' law

$F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}} F = \frac{1}{4 π \in_{0}} \frac{{qq}_{0}}{r^{2}} (q_{0} = 1 c) ∴ F \frac{1}{4 π \in_{0}} \frac{q}{r^{2}}$

This force acting on unit + ve charge is called intensity

So $[E = \frac{1}{4 π \in_{0}} \frac{q}{r^{2}}]$

Expression for electric potential at a point due to point charge $\to$

By coulombs' law

$F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{x^{2}} F = \frac{1}{4 π \in_{0}} \frac{q q_{0}}{x^{2}} (q_{0} = 1 c)$

Total work done is

$W = \int_{\propto}^{r} - Fdx = \int_{\propto}^{r} - \frac{1}{4 π \in_{0}} \frac{q}{x^{2}} dx = \frac{- q}{4 π \in_{0}} \int_{\infty}^{r} x^{- 2} dx = \frac{- q}{4 π \in_{0}} {[\frac{x^{- 1}}{- 1}]}^{r}_{\infty} W = + \frac{q}{4 π \in_{0}} {[\frac{1}{x}]}^{r}_{\infty} = \frac{q}{4 π \in_{0}} [\frac{1}{r} - \frac{1}{\infty}] ∴ W = \frac{1}{4 π \in_{0}} \frac{q}{r} (\frac{1}{\propto} = 0) ∴ W = v so [v = \frac{1}{4 π \in_{0}} \frac{q}{r}]$

To show the electric field is gradient of electric potential $\to$

The small work done in displacing $q_{0}$ from $P$ to $P^{1}$ is

$dw = - Fdr = - q_{0} Edr or \frac{dw}{q_{0}} = - Edr or dv = - Edr [E = - \frac{dv}{dr}]$

i.e., the electric intensity is the gradient of electric potential.

Define electric potential and obtain an expression between intensity of electric field due to a point charge and electric potential. Also show that electric field is gradient of electric potential.

Important Questions on Electric Potential

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