Question

Derive the relation between electric intensity and electric potential. 

Answer 1

Consider a point charge + q at point 'O'. The points M and N are separated by distance dx. The points Mand N are separated by distance dx.

Let q₀ be the test charge at point M. When it is displaced from M to N small work is done and it is given by

dω = −Fdx    ...(i)

Negative sign indicates force and displacement are in opposite direction.

By definition of electric intensity

`E=F/q_0`

Equation (i) becomes

`domega = -Eq_odx`

`(domega)/q_0 = -Edx`

`dV = -Edx, "where"  dV= (domega)/q_0`

`E=-(dV)/dx`

Thus the electric intensity at a point in an electric field is the negative potential gradient at that point.