Question

Two 2.00 mC point charges are located on the x-axis. One is at x=1.00 (a) Determine the electric field on the y-axis at y=0.500 m (b) Calculate the electric force on a -3.00C charge placed on the y-axis at y=0.500 m

Answer 1

A) find the electric field strength of the first charge at the point $y=0.5$ m: $E_1=k\frac{q_1}{r^2}=9\times 10^9\times \frac{2\times 10^{-3}}{0.5^2+1}=144\times 10^5$ V. According to the parallelogram rule, the sum of two equal stress vectors is

$E_{total}=\sqrt{E_1^2+E_2^2}=$$\sqrt{(144\times 10^5 )^2+(144\times 10^5 )^2}\approx 2.04 \times 10^7$ V.

B) since the electric field strength is $E=\frac{F}{q}$ , where $F$ is the Coulomb force, the force acting on the charge $a$ is $F=E_{total}\times q_a=2.04 \times 10^7\times 3=6.11\times 10^7$ N.