Question

Prove that electric field intensity is numerically equal to the negative of potential gradient.

Answer 1

NAZUAK❤

Answer. If you move a positive charge Q through this potential difference, the work done =Q x delta V where delta V is the change in potential/PD. You can also calculate the work done by using force x distance. ... E==delta V/ delta x ie minus the potential gradient.

Answer 2

Proved that electric field intensity is equal to the negative of potential gradient ⇒ $E = - \frac{dV}{dr}$.

Step-by-step Explanation:

To Prove: $E = - \frac{dV}{dr}$

Solution:

• Proof of $E = - \frac{dV}{dr}$

Consider a charge with the electric field intensity 'E' has displaced a small distance 'dr' such that it has done work against the electric field.

Therefore, by the definition of the potential difference 'dV', we have,

$dV = \vec{E} \cdot \vec{dr}$

$\Rightarrow dV = (E)(dr) cos \theta$

Since θ = 180° for the following case, therefore,

$\Rightarrow dV = - E \ dr$

$\Rightarrow E = - \frac{dV}{dr}$

Hence, proved that electric field intensity is equal to the negative of potential gradient ⇒ $E = - \frac{dV}{dr}$.