Question

Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm apart, each sphere is repelled from the other by a force of magnitude 2.0 N. Determine the magnitude of the smaller of the two charges.

Answer 1

Answer:

1.44 x 10⁻⁶ C

Explanation:

$q_{1}$ = charge on one sphere

$q_{2}$ = charge on other sphere

$q$ = Total charge on the two spheres = 40 μC

$q_{1}+ q_{2}$ = $q$

$q_{1}+ q_{2}$ = 40 x 10⁻⁶

$q_{1}$ = (40 x 10⁻⁶) - $q_{2}$                                   eq-1

$r$ = distance between the two spheres = 50 cm = 0.50 m

$F$ = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

$F = \frac{k q_{1} q_{2}}{r^{2}}$

$2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}$

$q_{2}$ = 1.44 x 10⁻⁶ C