Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin

(a) must be equal to 20 Vcm^{−1}

(b) may be equal to 20 Vcm^{−1}

(c) may be greater than 20 Vcm^{−1}

(d) may be less than 20 Vcm^{−1 }

#### Solution

(b) may be equal to 20 Vcm^{−1}

(c) may be greater than 20 Vcm^{−1}

Change in the electric potential, dV = 40 V

Change in length, \[∆ r\] = −1−1 = −2 cm

Electric field, \[E = \frac{- dV}{dr}\]

\[\Rightarrow E = - \frac{40 V}{- 2}\]

\[ \Rightarrow E = 20 {Vcm}^{- 1}\]

This is the value of the electric field along the x axis.

Electric field is maximum along the direction in which the potential decreases at the maximum rate. But here, direction in which the potential decreases at the maximum rate may or may not be along the x-axis. From the given information,the direction of maximum decrease in potential cannot be found out accurately. So, E can be greater than 20 V/cm in the direction of maximum decrease in potential.

So, the electric field at the origin may be equal to or greater than 20 Vcm^{−1}.