An infinite number of charges of equal magnitude q, but of opposite sign are placed alternately starting with positive charge along x- axis at x = 1 m; x = 2 m,x = 4 m, x= 8 m and so on. The electric potential at the point x = 0 due to these charges will be (in S.I units and k = $\frac{1}{4 π ϵ_{0}}$

\frac{k q}{2}

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\frac{k q}{3}

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\frac{k 2 q}{3}

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\frac{k 3 q}{2}

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Solution

The correct option is C $\frac{k 2 q}{3}$
$V = V_{1} + V_{2} + V_{3} + . . . . . . . = \frac{1}{4 π ϵ_{0}} [\frac{q_{1}}{r_{1}} - \frac{q_{2}}{r_{2}} + \frac{q_{3}}{r_{3}} - \frac{q_{4}}{r_{4}} + . . . . . . .] = k [\frac{q}{1} - \frac{q}{2} + \frac{q}{4} - \frac{q}{8} + . . . . . .] = k q [1 - \frac{1}{2} + \frac{1}{2^{2}} - \frac{1}{2^{3}} + . . . . . .] = k q [\frac{1}{1 + \frac{1}{2}}] = \frac{2 k q}{3}$

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q

and of the same sign, are placed along the x-axis at

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x = 0

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\frac{1}{4 π ϵ_{0}} = k

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and of the same sign, are placed along the x-axis at

x = 1 m, x = 2 m, x = 4 m, x = 8 m

and so on. Then the electrical potential at

x = 0

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\frac{1}{4 π ϵ_{0}} = k

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An infinite number of charges of equal magnitude q, but of opposite sign are placed alternately starting with positive charge along x- axis at x = 1 m; x = 2 m,x = 4 m, x= 8 m and so on. The electric potential at the point x = 0 due to these charges will be (in S.I units and k = 14πϵ0

The correct option is C k2q3V=V1+V2+V3+.......=14πϵ0[q1r1−q2r2+q3r3−q4r4+.......]=k[q1−q2+q4−q8+......]=kq[1−12+122−123+......]=kq[11+12]=2kq3