Question

relationship between electric field and electric potential​

Answer 1

Answer:

E = V / R

Explanation:

E=kq/r^2

V=kq/r

K=1/4π€

Answer 2

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To derive :-

Relationship between Electric Field and Electric Potential .

Proof:-

First of all , let's consider a charge q located at a specified point and emitting electric field . We shall also consider 2 closely located points named as A and B.

The distance between A and B be dx.

Since the dx distance is extremely small, we can consider that the Electric field E (from the charge) remains same at both the points.

Now the work done to move a unit positive charge from A to B will be :

$\large\dashrightarrow{ \tt{dW = - E \times dx }}$

The negative sign denoted that the work done was against the Electric Field Direction.

For a unit positive charge , we know that work done us equal to the potential difference dV.

$\large\dashrightarrow \: \: \tt{dV = - E \times dx }$

$\large{ \tt{ \dashrightarrow E = - \dfrac{dV}{dx} }}$

So we can say that Electric Field is the gradient of Electric Potential with respect to distance.

$\huge{\mathfrak{\orange{hope\; it\; helps}}}$