Question

What will be the value of both charges if they are 5 cm apart and suffer a
attraction force of 5.2 N.

Answer 1

Answer:

$\boxed{q = 1.2 \times {10}^{ - 6} C}$

Explanation:

$f_e = \frac{{q}^{2}k }{ {r}^{2} } \\ q = \sqrt{ \frac{f_e( {r}^{2} )}{k} } = \sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } \\ q =\sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } = \sqrt{ \frac{0.013}{9 \times {10}^{9} } } \\ q = 1.2 \times {10}^{ - 6}$