Question

Infinite charges are lying at x = 1, 2, 4, 8…meter on X-axis and the value of each charge is Q. The value of intensity of electric field and potential at point x = 0 due to these charges will be respectively

• A. $12 \times 10^{9}Q$N/C, 1.8 × 10⁴
• B. Zero, 1.2 × 10⁴V
• C. $6 \times 10^{9}Q$N/C, 9 × 10³V
• D. $4 \times 10^{9}Q$N/*C,*6 × 10³V

Answer 1

Answer: (A)

$12 \times 10^{9}Q$N/C, 1.8 × 10⁴

Answer 2

By the superposition, Net electric field at origin

$E = kQ\left\lbrack \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{8^{2}} + ...\infty \right\rbrack E = kQ\left\lbrack 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...\infty \right\rbrack$

$1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...\infty$ is an infinite geometrical progression it’s sum can be obtained by using the formula $S_{\infty} = \frac{a}{1 - r}$ ; Where a = First term, r = Common ratio.

Here $a = 1$ and $r = \frac{1}{4}$ so,

$1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + .....\infty = \frac{1}{1 - 1/4} = \frac{4}{3}$.

Hence $E = 9 \times 10^{9} \times Q \times \frac{4}{3} = 12 \times 10^{9}QN ⥂ / ⥂ C$Electric

potential at origin

$V = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{1 \times 10^{- 6}}{1} + \frac{1 \times 10^{- 6}}{2} + \frac{1 \times 10^{- 6}}{4} + \frac{1 \times 10^{- 6}}{8} + .......\infty \right\rbrack = 9 \times 10^{9} \times 10^{- 6}\left\lbrack 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ............\infty \right\rbrack = 9 \times 10^{3}\left\lbrack \frac{1}{1 - \frac{1}{2}} \right\rbrack = 1.8 \times 10^{4}volt$

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Note : In the arrangement shown in figure +Q and – Q are alternatively and equally spaced from each other, the net potential at the origin O is $V = \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \cdot \frac { Q \log _ { e } 2 } { x }$