Question

Two small conducting point charges, separated by 0.6 m, carry a total charge of 185 mu or micro CC. They repel one another with a force of 120 N. For the universal constant k use the value 8.99 times 109 N m2/C2. Find the charge on the larger of the two point charges:

Answer 1

Answer:

Explanation:

Let the two charges are q and Q.

According to the question

q + Q = 185 micro Coulomb .... (1)

Force of repulsion, F = 120 N

distance, d = 0.6 m

K = 8.99 x 10^9 Nm²/C²

Use Coulomb's law

$F = \frac{kQq}{d^{2}}$

$120= \frac{8.99\times 10^{9}Q\left ( 185-Q \right )\times 10^{-12}}{0.6^{2}}$    .... from equation (1)

4805.4 = 185 Q - Q²

Q² - 185 Q + 4805.4 = 0

$Q=\frac{185\pm \sqrt{185^{2}-4\times 4805.4}}{2}$

Q = 153.75 μC, 31.25 μC

q = 185 - 153.75 = 31.25 μC, 185 - 31.25 = 153.75 μC

So, charges are

31.25 μC, 153.75 μC