Question

Two charges are placed on the x-axis. One of the charges (q1 = +9.0 µC) is at x1 = +3.2 cm and the other (q2 = -24 µC) is at x2 = +8.1 cm.
(a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the x-axis.)
(b) Find the net electric field (magnitude and direction) at x = +5.7 cm. (Use the sign of your answer to indicate the direction along the x-axis.)

Answer 1

The electric field of a point charge

$$E=k\frac{q}{r^2}.$$

(a)

The net field

$$E=E_2-E_1=k\frac{q_2}{r_2^2}-k\frac{q_1}{r_2^1},$$

$$=9\times 10^9\left(\frac{24\times 10^{-9}}{(0.081)^2}-\frac{9\times 10^{-9}}{(0.032)^2}\right)=-4.62\times 10^4\:\rm V/m$$

(b)

The net field

$$E=E_2+E_1=k\frac{q_2}{r_2^2}+k\frac{q_1}{r_2^1},$$

$$=9\times 10^9\left(\frac{24\times 10^{-9}}{(0.024)^2}+\frac{9\times 10^{-9}}{(0.024)^2}\right)=5.16\times 10^5\:\rm V/m$$