A charge +q is fixed at each of points x = x_0,,x = 3x_0,,x = 5x

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2. Electric Potential and Capacitance

medium

A charge $+q$ is fixed at each of the points $x = x_0,\,x = 3x_0,\,x = 5x_0$, .... upto $\infty $ on $X-$ axis and charge $-q$ is fixed on each of the points $x = 2x_0,\,x = 4x_0,\,x = 6x_0$, .... upto $\infty $ . Here $x_0$ is a positive constant. Take the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{{4\pi {\varepsilon _0}r}}$. Then the potential at the origin due to above system of charges will be

zero

$\frac{q}{{8\pi {\varepsilon _0}{x_0}\,{{\log }_e}\,2}}$

infinity

$\frac{{q\,{{\log }_e}\,2}}{{4\pi {\varepsilon _0}{x_0}}}$

Solution

Potential at origin $=\left(\mathrm{V}_{1}+\mathrm{V}_{3}+\mathrm{V}_{5}+\ldots \ldots\right)-\left(\mathrm{V}_{2}+\mathrm{V}_{4}+\mathrm{V}_{6}+\ldots .\right)$

$\Rightarrow \frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{\mathrm{x}_{0}}-\frac{1}{2 \mathrm{x}_{0}}+\frac{1}{3 \mathrm{x}_{0}} \ldots \infty\right]$

$\Rightarrow \frac{q}{4 \pi \varepsilon_{0} x_{0}}\left[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \ldots . \infty\right]$

$\Rightarrow \frac{q}{4 \pi \varepsilon_{0} \mathrm{x}_{0}} \log _{\mathrm{e}}(1+1) \Rightarrow \frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{x}_{0}} \log _{\mathrm{e}} 2$

Standard 12

Physics

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easy

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hard

(JEE MAIN-2017)

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medium

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easy

(NEET-2022)

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easy

Solution

Similar Questions

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