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serg [7]
2 years ago
9

What will be the value of both charges if they are 5 cm apart and suffer a

Physics
1 answer:
Mrac [35]2 years ago
3 0

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}

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