Two charges `+ 10 muC` and `+ 20 muC` are placed at a. separation of 2 cm. Find the electric potential due to the. pair at the middle point of the line joining the two charges.

Explanation

Using the equation `V = Q/4 pi epsilon_0 r, the potential due to
`+ due muC` is
`V_1 = (( 10xx10(-6) C)) x ((9xx10^9 Nm^2 C(-2) ))/(1x10(-2)m) = 9MV.`
The potential due to `+20 muC is v_2 +((20 xx 10 (-6) C)) xx ((9xx10^ 9 Nm ^ 2 C(-2) )) /1x10 9-2) m = 18 MV`
The net potential at the given point is
9 MV + 18 MV = 27 MV.
If the charge distribution is continuous, we may
use the technique of intergration to find the electric potential.