Ten positively charged particles are kept fixed on x – axis at points x = 10cm, 20 cm, 30cm,….100 cm,. First particle has a charge $10^{- 8} C$ , the second $8 \times 10^{- 8} C$ the third $27 \times 10^{- 8} C$ and so on. The tenth particle has a charge $1000 \times 10^{- 8} C$ . The magnitude of electric force acting on a 1 C charge placed at origin is

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5.94 \times 10^{5} N

5 .49 \times 10^{5} N

4.95 \times 10^{5} N

9.45 \times 10^{5} N

answer is D.

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Detailed Solution

$F_{nt} = \frac{1}{4 π ε_{0}} (\frac{{qq}_{1}}{r_{1}^{2}} + \frac{{qq}_{2}}{r_{2}^{2}} + \frac{{qq}_{3}}{r_{3}^{2}} + \dots \dots + \frac{{qq}_{10}}{r_{10}^{2}})$

$= \frac{q}{4 π ε_{0}} [\frac{q_{1}}{r_{1}^{2}} + \frac{q_{2}}{r_{2}^{2}} + \frac{q_{3}}{r_{3}^{2}} + \dots \dots + \frac{q_{10}}{r_{10}^{2}}]$

$= 1 \times 9 \times 10^{9} [\frac{10^{- 8}}{{(10 \times 10^{- 2})}^{2}} + \frac{8 \times 10^{- 8}}{{(20 \times 10^{- 2})}^{2}} + \frac{27 \times 10^{- 8}}{{(30 \times 10^{- 2})}^{2}} + \dots \dots]$

$= \frac{9 \times 10^{9} \times 10^{- 8}}{10^{- 4}} [\frac{1}{10^{2}} + \frac{8}{20^{2}} + \frac{27}{30^{2}} + \dots \dots + \frac{10^{3}}{{(100)}^{2}}]$