Question

1. (15 points) Three charges are shown below. Find the following values: a. What is the electric field at the point labeled A? b. What is the electric potential at the point labeled A? c. What is the electric force on the +10µC charge?

—6л с 4 thuc 0.3 м 0.3 м 0. Чим tlouc

Answer 1

a. The E field at point A can be calculated using the superposition principle, adding up the contributions of the 3 charges:

N E-64c = 9 x 109Nm? 6 x 10-60 (0.3m)2 600,000

Since the E field of a negative charge points radially inwards and point A is at its right we can write

$\bar{E}_{-6\mu C}=-600,000\hat{i}\frac{N}{C}$

The +6uC charge is located at the same distance so the E contribution will have the same magnitude, however for a positive point charge the field points radially outwards and point A is at its left

$\bar{E}_{6\mu C}=-600,000\hat{i}\frac{N}{C}$

Then for the 10uC charge we have

E104C = 9 x 109 Nm2 X 10 x 10-60 (0.4m2 N 562, 500-

And the direction:

$\bar{E}_{10\mu C}=562,500\hat{j}\frac{N}{C}$

So the total electric field at point A is

$\bar{E}_{A}=(-1,200,000\hat{i}+562,500\hat{j})\frac{N}{C}$

$E_{A}=\sqrt{(-1,200,000\frac{N}{C})^{2}+(562,500\frac{N}{C})^{2}}=1,325,295\frac{N}{C}$

b. The electric potential at point A is given by:

$V_{A}=9\times 10^{9}\frac{Nm^{2}}{C^{2}}\left ( \frac{-6\times 10^{-6}C}{0.3m}+\frac{6\times 10^{-6}C}{0.3m}+\frac{10\times 10^{-6}C}{0.4m} \right )$

$V_{A}=225,000V$

c. The 10 uC charge is at the same distance d = sqrt((0.3m)^2 + (0.4m)^2) = 0.5m from the other 2 charges, and it's attracted to the negative and repeled by the positive. Because of the simmetry the vertical contributions cancel each other and the horizontal contributions add up to give

$\bar{F}_{10\mu C}=\left (-\frac{2\times 9\times 10^{9}\frac{Nm^{2}}{C^{2}}\times 6\times 10^{-6}C\times 10\times 10^{-6}C}{(0.5m)^{2}}\times \frac{0.3m}{0.5m} \right )\hat{i}$

$\bar{F}_{10\mu C}=-2.592\hat{i}N$