Question

Two charges are placed on the x axis. One of the charges (q1 = +7.86 uC) 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

Electric field at x=0 due to due to charge $q1$ is given by

$E_1=k\frac{q_1}{(x_1-x)^{2}}=\frac{7.86\times10^{-6}\times8.99\times 10^{9}}{0.03^2}=7.9\times10^7N/C$

because the charge is positive it acts a long the negative x direction

Similarly electric field at x=0 due to charge $q_2$ is given by

$E_2=k\frac{q_2}{(x_2-x)^{2}}=\frac{22.3\times10^{-6}\times8.99\times 10^{9}}{0.09^2}=2.5\times10^7N/C$

since the charge is negative it acts along the positive x direction

so the net electric field is given by;

$E_{net}=7.9\times10^7-2.5\times10^7=5.4\times10^7 N/C$

Electric field at x=+0.06m due to due to charge $q1$ is given by

$E_1=k\frac{q_1}{(x-x_1)^{2}}=\frac{7.86\times10^{-6}\times8.99\times 10^{9}}{(0.06-0.03)^2}=7.9\times10^7N/C$

its along the positive x direction

Electric field at x=+0.06m due to due to charge 2 is given by

$E_2=k\frac{q_2}{(x_2-x)^{2}}=\frac{22.3\times10^{-6}\times8.99\times 10^{9}}{(0.09-0.06)^2}=2.23\times10^8N/C$

this also acts along the positive x direction

net electric field along at x=+0.06m is

$E_{net}=7.9\times10^7+2.23\times10^8=3.02\times10^8N/C$