Question

The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.

Answer 1

The electric field value is 240 N/C

Step-by-step explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

`\Delta V=Ed`

`E=(\Delta V)/(d)`

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

`E=(1.2)/(5.0*10^(-3))`

`E= 240\ N/C`

Hence, The electric field value is 240 N/C