Question

Two charges are placed on the x-axis. One charge (q1 =+8.5 µC) is at x1 = +3.0 cm and the other (q2 = -21 µC) is at x2 = +9.0 cm. Find the net electric field (magnitude and direction) at x = +6.0 cm.

Answer 1

Both field will direct towards right as positive charge has electric field away from the charge and the field will point inside the charge due to negative charge resulting in net field towards right in between of the two point charges.

Electric field due to both charges$(E) = k\frac {q_1}{r_1^2} + k\frac{q_2}{r_2^2} = 9.0\times10^9\times$ $(\frac{8.5\times10^{-6}}{0.030^2} +$ $\frac{21\times10^{-6}}{0.030^2} )= 2.94\times 10^8\frac{N}{C}$ $\ \hat{i}$