Answer to Question #190161 in Electric Circuits for D.Weerakkodi

Question

Points A and B are 0.10 m apart. A point charge of +3.0 x 10-9 C is placed at A and a

point charge of -1.0 x 10-9 C is placed at B.

I. X is the point on the straight line through A and B, between A and B, where the

electric potential is zero. Calculate the distance AX.

II. Show on a diagram the approximate position of a point, Y, on the straight line

through A and B where the electric field strength is zero.( no calculation is

expected)

Answer 1

Charge at A, $q_1=+3.0\times10^{-9}\space C$

Charge at B, $q_2=-1\times10^{-9}\space C$

Potential at X due to charge $q_1$

$V_1=\dfrac{kq_1}{x}$

Potential at X due to charge $q_2$

$V_2=\dfrac{kq_2}{(0.10-x)}$

Total electric potential at X is equal to 0

$V_1+V_2=0$

$\dfrac{k(3\times10^{-9})}{x}-\dfrac{k(1\times10^{-9})}{(0.10-x)}=0$

$4x=0.30$

$x=0.075\space m$

(II)