Question

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Ten Positively charged particles are kept fixed on x - axis at points x = 10 cm , 20 cm , 30 cm ..... 100 cm . The first particle has a charge of $1 \times 10^{-8} \: C$ The second of $8 \times 10^{-8} \: C$ , The third of $27 \times 10^{-8} \: C$ and so on . The tenth particles has a charge of $1000 \times 10^{-8} \: C$.

Then Find the magnitude of the electric force acting on a 1C charge placed at the origin .


→ Explanation required


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Answer 1

Given :- 10 particles are place on x axis at a equidistance of 10 cm and their charges are given .

Question :- To find magnitude of the electric force acting on a 1C charge placed at the origin .

Solution :-

we have to find ' f '

» if suppose Three particles are given then to find 'F ' we add F1 + F2 + F3..

» Similarly for 10 particles

F = F1 + F2 + F3...........+ F10

By columbs law we know that :-

F = k. q1 .q2/r^2

Given q2 is the particle placed at the origin with 1C as it's charge.

k = 9 × 10 ^9

r1 = 10 cm = 10 × 10 ^ -2

similarly convert all the "r " into meters .and remember that r2 = 20 cm , r3 = 30.....

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Solution is in the attachment , kindly refer it.

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