Question

Derive
the relation between electric field and electric potential
​

Answer 1

Explanation:

Consider two points A and B separated by a small distance dx in an electric field.

Since dx is small, the electric field E is assumed to be uniform along AB. The force acting on a unit positive charge at A is equal to E.

Now, the work done in moving a unit positive charge from A to B against the electric field is dW = − E× dx.

The negative sign shows that the work is done against the direction of the field. Since the work done is equal to potential difference dV between A and B then 

dV = −E × dx or E = - dv/dx

Thus, the electric field at a point is the negative of the potential gradient at that point.

Hope its helpful for you....✌✌

Answer 2

Answer:-

Consider two points A and B separated by a small distance dx in an electric field. Since dx is small, the electric field E is assumed to be uniform along AB. The force acting on a unit positive charge at A is equal to E .

Now, the work done in moving a unit positive charge from A to B against the electric field is

$\rm \: \: \: \: \: \: \: \: \: \: \: dW = -E \times dx$

The negative sign show that the work is done against the direction of the field.

Since, the work done is equal to the potential difference dV between A and B then

$\\ \: \: \: \: \: \: \: \: \: \: \rightarrow\rm \: dV = - E \: \: or \: \: E \: = \: - \frac{dV}{dx}$

Thus, the electric field at a point is the negative of the potential gradient at that point.