Question

The force of attraction between the charges 0.4mC and -0.2mC is 2N. Find the distance between them.

Answer 1

Answer :-

$\implies \boxed{\sf{r = 6 \times {10}^{ - 3} \: m \: } } \\ \bf{ or }\: \\ \implies \boxed{\sf{ r = 6 \: millimetre \: }} \:$

To Find :-

→ Distance between two given charges .

Explanation :-

Given that :-

• Force of attraction between charges (F) = 2 N

• $\sf{q_{1} = 0.4 \: \mu \: c \: }$

• $\sf{q_{2} = 0.2\: \mu \: c \: }$

• Distance between them (r) = ?

We know that

→ force between two charge

$\implies \: \boxed{ \sf{F = \frac{1}{4 \pi\epsilon_{o} } \: \frac{q_{1} q_{2} }{ {r}^{2} } }}$

therefore ,

$\because \sf{1 \mu = {10}^{ - 6} } \: \\ \therefore \: \\ \implies \sf{2 = \frac{1}{4 \pi\epsilon_{o} } \frac{0.2 \times 0.4 \times {10}^{ - 12} }{ {r}^{2} } } \\ \\ \because \boxed{ \sf{\frac{1}{4 \pi\epsilon_{o} } = 9 \times {10}^{9} }} \\ \\ \implies \sf{ 2 = 9 \times {10}^{9} \: \times \frac{8 \times {10}^{ - 14} }{ {r}^{2} } } \\ \\ \implies \sf{2 \: {r}^{2} = 72 \times {10}^{ - 5} } \\ \\ \implies \sf{ {r}^{2} = 36 \times {10}^{ - 5} } \\ \\ \implies \sf{ {r}^{2} = 3.6 \times {10}^{ - 4} }$

$\implies \sf{r = 0.6 \times {10}^{ - 2} } \\ \: \\ \implies \boxed{\sf{r = 6 \times {10}^{ - 3} \: m \: } } \\ \bf{ or }\: \\ \implies \boxed{\sf{ r = 6 \: millimetre \: }}$

Hence , distance between the given charges is 6 millimetre .