a) Let x be measured to the right of the 3uC charge, which we will assume to be the left-hand charge. For zero field strength,
k*q1/x^2 + kq2/(x-5)^2 = 0
[(x-5)/x]^2 = -q2/q1 = 2/3
(x-5)/x = 0.8165 = 1 - (5/x)
5/x = 0.1835
x = 27.2 cm to the right of q1 and, which is 22.2 cm to the right of q2.
b) For zero potential,
-k*q1/x -k*q2/(x-5) = 0
(x-5)/x = -q2/q1 = 2/3
1 - (5/x) = 2/3
5/x = 1/3
x = 15 cm
two point charges q1=3uc and q2=-2uc are placed 5 cm apart on the x asix. at what points along the x asix in
a) the electric field zero and
b) the potential zero ?
2 answers
You should add 1 to 2/3
5 answers