Question

Two charges are placed on the x axis. One of the charges (q1 = +7.86C) is at x1 = +3.00 cm and the other (q2 = -22.3C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

Answer 1

(a) the projection of field on the x-axis from the first charge is

$E_{1x}=-k\dfrac{q_1} {x_1^2} ,$ from the second charge $E_{2x}=k\dfrac{|q_2|} {x_2^2} ,$ so the net field is $E=-k\dfrac{q_1} {x_1^2} + k\dfrac{|q_2|} {x_2^2} = - 9\cdot10^9\cdot \dfrac{7.86} {0.03^2} + 9\cdot10^9\cdot \dfrac{22.3} {0.09^2} = - 5.4\cdot10^{13}\text{N/C}.$

The charges are quite large, so the field is also large.

(b) the projection of field on the x-axis from the first charge is

$E_{1x}=k\dfrac{q_1} {(x-x_1)^2 } ,$ from the second charge $E_{2x}=k\dfrac{|q_2|} {(x_2-x) ^2} ,$ so the net field is $E=k\dfrac{q_1} {(x-x_1) ^2} + k\dfrac{|q_2|} {(x-x_2) ^2} = 9\cdot10^9\cdot \dfrac{7.86} {(0.06-0.03)^2} + 9\cdot10^9\cdot \dfrac{22.3} {(0.06-0.09)^2} = 3\cdot10^{14}\text{N/C}.$