# Relation between electric fields and Electric potential

## Relation between electric fields and electric potential

- Consider the electric field E due to a point charge +q at point O in a radially outward direction shown below in the figure.

- Suppose R and S are two points at a distance r and r+dr from point O where dr is vanishingly small distance and V is electric potential at point R.

- Now force on any test charge q' at point R in terms of electric field is

**F**=q'**E**

- Work done by the force in displacing test charge from R to S in field of charge q is

dW = **F**·d**r** = q'**E**·d**r**

and, change in potential energy is

dU = -dW = -q'**E**·d**r**

Change in electric potential would be

dV = dU/q

or dV = -**E**·d**r** ----(11)

- From equation 11 electric field is

$E=- \frac {dV}{dr}$ ---(12)

The quantity $\frac {dV}{dr}$ is the rate of change of potential with the distance and is known as **potential gradient**. Negative sign in equation 12 indicates the decrease in electric potential in the direction of electric field.

- For Cartesian coordinate system

**E**=E_{x}**i**+E_{y}**j**+E_{z}**k**

and,

d**r**=dx**i**+dy**j**+dz**k**

from equation 11

dV=-**E** . d**r**

or, $dV=-(E_x dx+E_y dy+E_zdz)$ ----(13)

- Thus components of
**E** are related to corresponding derivatives of V in the following manner

$E_x=-\frac {dV}{dx}$ ---(14a)

$E_y=-\frac {dV}{dy}$ ---(14b)

$E_z=-\frac {dV}{dz}$ -----(14c)

In equation (14a) we see that V is differentiated with respect to coordinate x keeping other coordinates constant. Same is the case with equations (14b) and (14c) in case of y and z coordinates respectively.

- This can be written in partial derivative form as
$E_x=-\frac{\partial V}{\partial x}$ ---(15a)

$E_y=-\frac{\partial V}{\partial y}$ ---(15b)

$E_z=-\frac{\partial V}{\partial z}$ -----(15c)

- So if we know Electric Field as function of position we can calculate V using Eq. 11, and if we know V as a function of position,
we can calculate using Eq. 14 Deriving V from requires integration, and deriving from V requires differentiation.

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