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Physics Question on electrostatic potential and capacitance

An infinite number of charges each of magnitude q are placed-on x - axis :at distances of 1,2, 4, 8, ... meter from the origin. The intensity of the electric field at origin is

A

q3πε0\frac{q}{3\pi\varepsilon_0}

B

q6πε0\frac{q}{6\pi\varepsilon_0}

C

q2πε0\frac{q}{2\pi\varepsilon_0}

D

q4πε0\frac{q}{4\pi\varepsilon_0}

Answer

q3πε0\frac{q}{3\pi\varepsilon_0}

Explanation

Solution

We know E=kqr2E =\frac{ kq }{ r ^{2}} Now electric field at 0=kq12+kq22+kq42+kq82+−−−−0=\frac{ kq }{1^{2}}+\frac{ kq }{2^{2}}+\frac{ kq }{4^{2}}+\frac{ kq }{8^{2}}+---- =kq(1+122+124+126+−−−−)= kq \left(1+\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+----\right) We know in G. P sum to infinite =a1−r=\frac{ a }{1- r } =kq⋅(11−14)= kq \cdot\left(\frac{1}{1-\frac{1}{4}}\right) =q4πε0⋅43=q3πε0=\frac{ q }{4 \pi \varepsilon_{0}} \cdot \frac{4}{3}=\frac{ q }{3 \pi \varepsilon_{0} }