Question

Infinite charges of magnitude $q$ each are lying at $ x=1,2,4,8,... $ metre on $X$-axis. The value of intensity of electric field at point $ x=0 $ due to these charges will be

• A. $ 12\times 10^{9}q\,N/C $
• B. zero
• C. $ 6\times 10^{9}q\,N/C $
• D. $ 4\times 10^{9}q\,N/C$

Answer 1

Answer: (A)

If there are $n$ point charges $q_{1},\, q_{2}, \ldots \ldots q_{n}$, then each of them will produce the same intensity at any point which it would have produced in the absence of other point charges.
Hence, total intensity will be vector sum of $\vec{E_{1}}, \vec{E_{2}} \ldots, \vec{E_{n}}$ produced at a point.
$\Sigma E=\frac{q}{4 \pi \varepsilon_{0} r_{1}^{2}}+\frac{q}{4 \pi \varepsilon_{0} r_{2}^{2}}+\ldots . .+\frac{q}{4 \pi \varepsilon_{0} r_{n}^{2}}$
$\therefore \vec{E}=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{8^{2}}+\ldots . .\right]$
The given series is a geometric progression.
Hence, sum $(S)=\frac{a}{1-r}$ where a is first term of series and $r$ the common difference.
$r=\frac{1}{4},\, a=1$
$\therefore S=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
$\therefore $ Total intensity $=\frac{q \times 4}{4 \pi \varepsilon_{0} \times 3}=\frac{9 \times 10^{9} \times 4 q}{3}$
$=12 \times 10^{9} q N / C$