Question

8. Two positive charges (+8.0 mC and +2.0 mC) are separated by 300 m. A third charge is placed at distance r from the +8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is

Answer 1

Answer:

0.20 km

Explanation:

The computation of the distance r is shown below:

This can be calculated by using the following formula

As we know that

$\frac{k_q_1}{r^2} = \frac{k_q_1}{(300 - r)^2}$

where,

q_1 is 8 mC

q_2 is 2 mC

Now placing these values

Now

$\frac{k (8 \times 10^{-3c})}{r^2} = \frac{k (2 \times 10^{-3c})}{(300 - r)^2}$

$\frac{m}{r^2} = \frac{1}{(300 - r)^2}$

$\frac{2}{r} = \frac{1}{300 - r}$

r = 600 - 2r

r = 200 m

r = 0.20 km

We simply applied the above formula to determine the distance of r  

Hence, the distance r is 0.20 km