**Question: Establish relation between electric field intensity and potential gradient.**

**Ans:**Let A and B are two points on a line of force due to a point charge +Q, as shown in the figure. These points are so close that electric field intensity (E) is constant between them.

Work done in taking a unit charge from B to A = Force * displacement

= - E dx ….... (i)

(.: E = F/q

^{|}for unit charge q^{|}= 1 .: E = F)Negative sign shows that force on the unit charge and displacement are in opposite directions.

If the potential difference between A and B is dV, (A is at higher potential), then work done in taking a unit charge from B to A = dV …..... (ii)

From (i) and (ii), we get

dV = -E dx

E = dV/dx …….. (iii)

Thus, electric field strength is equal to the negative of the potential gradient. The direction of E is always in the direction of decrease of potential.

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