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

`\boxed{q = 1.2 * {10}^( - 6) C}`

Step-by-step explanation:

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