Question

Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 46.0 N on the other, find the magnitudes of the charges

Answer 1

Answer:

The charge on point A is $q_a$ = 2.4 × $10^{-5}$ C

The charge on point B is $q_b$ = 1.2 × $10^{-5}$ C

Explanation:

Given data

Distance (r) = 0.19 m

Magnitude of the charge on A is twice that of the charge on B  i.e.

$q_A = 2 q_B$

F = 46 N

We know that force between the charges is given by

$F = \frac{k Q_A Q_B}{r^{2} }$

$65 = \frac{(9) (10^{9}) (2q_b^{2} ) )}{0.2^{2} }$

$q_b$ = 1.2 × $10^{-5}$ C

$q_a$ = 2 × $q_b$

$q_a$ = 2 × 1.2 × $10^{-5}$

$q_a$ = 2.4 × $10^{-5}$ C

Therefore the charge on point A is $q_a$ = 2.4 × $10^{-5}$ C

The charge on point B is $q_b$ = 1.2 × $10^{-5}$ C