A -3.7 nC charge is on the x-axis at X1= -7 cm and a 4.2 nC charge is on the x-axis at X2 = 20 cm.
If the potential at (0, u) is 0, so also is the potential at (0, -u).
Potential at (0, u) due to the -3.7nC charge is kQ/r where r is the distance between (0, u) and (-7, 0).
Here r = (u (-7)) = (u 49)
So this potential is (9.0 x 10 x (-3.7) x 10 )/(u 0.0049)
Similarly the potential due to the 4.2nC charge is
is (9.0 x 10 x 4.2 x 10 )/(u 0.04)
If the total potential at (0, u) is 0, then
-(9.0 x 10 x 3.7 x 10 )/(u 0.0049)
9.0 x 10 x 4.2 x 10 )/(u 0.04), giving
(u 0.0049)/3.7 = (u 0.04)/4.2
4.2(u 0.0049) = 3.7(u 0.04)
u = 0.34m = 34cm.
The points on the y-axis of zero potential are
(0, 34), (0, -34).
This post is last updated on hrtanswers.com at Date : 1st of September – 2022