Question

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Answer 1

Answer =

Let total Force = F

then F

=$\frac{k×1×2 × micro^2C^2}{1^2} +\frac{k×1×2 × micro^2C^2}{2^2}+ \frac{k×1×2× micro^2C^2}{4^2} + ...........$

= $2k× micro^2 C^2 ( 1 + \frac{1}{4} +\frac{1}{16} +\frac{1}{64} + ........)$

Which forms a G.P

And G.P

= $\frac{a}{1-r}$

and here
a = 1

and

r = $\frac{1}{4}$

Then value of GP

= $\frac{1}{1-{\frac{1}{4}}}$

=$\frac{1}{{\frac{4-1}{4}}}$

= $\frac{1}{{\frac{3}{4}}}$

= $\frac{4}{3}$

Then by putting value of GP

we get

= $2k × micro^2 C^2 × \frac{4}{3}$

= $2× 9 × 10^9 × micro^2 C^2 × \frac{4}{3}$

= $24 × 10^9 × 10^{-12} N$

= $24 × 10^{-3} N$

= $2.4 × 10^{-2} N$