Question

The electrostatic force between charges of 200 µC and 500 µC placed in free space is 5 gf. Find the distance between the two charges. Take g = 10 ms-2​

Answer 1

$\mathtt{ \huge{\fbox{Solution :)}}}$

Given ,

• The two charges are 200 µC or 200 × (10)^-6 and 500 µC or 500 × (10)^-6 C

• The electrostatic force between them is 5 gf or 5 × (10)^-2 newton

We know that , the electrostatic force between two charges is given by

$\huge{ \mathtt{ \fbox{Force = k\frac{ q_{1}q_{2}}{ {(r)}^{2} }} }}$

Thus ,

$\sf \mapsto 5 \times {(10)}^{ - 2} = \frac{9 \times {(10)}^{9} \times 200 \times {(10)}^{ - 6} \times 500 \times {(10)}^{ - 6} }{ {(r)}^{2} } \\ \\ \sf \mapsto 5 \times {(10)}^{ - 2} = \frac{9 \times {(10)}^{9} \times 2 \times {(10)}^{2} \times {(10)}^{ - 6} \times 5 \times {10)}^{2} \times {(10)}^{ - 6} }{ {(r)}^{2} } \\ \\ \sf \mapsto {(r)}^{2} = \frac{9 \times {(10)}^{9} \times 2 \times {(10)}^{2} \times {(10)}^{ - 6} \times 5 \times {(10)}^{2} \times {(10)}^{ - 6} }{ 5 \times {(10)}^{ - 2} } \\ \\ \sf \mapsto {(r)}^{2} = 18 \times {(10)}^{(9 + 2 - 6 + 2 - 6 + 2))} \\ \\ \sf \mapsto {(r)}^{2} = 18000 \\ \\ \sf \mapsto r = \sqrt{18000} \\ \\ \sf \mapsto r =134 \: \: m \: \: (approx)$

Hence , the distance between two charges is 134 m

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