An infinite number of charges, each numerically equal to $q$ 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 $x = 0$ due to this set of charges is ( Assume that $\frac{1}{4 π ϵ_{0}} = k$ )

2 k q

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\frac{2 k q}{5}

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\frac{4 k q}{3}

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\frac{4 k q}{5}

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Solution

The correct option is A $2 k q$
Potential at origin due to other charges
$V = \frac{k q}{1} + \frac{k q}{2} + \frac{k q}{4} + \frac{k q}{8} + . . . . .$
$V = k q [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . .]$
$V = k q ⎡ ⎢ ⎢ ⎢ ⎣ \frac{1}{1 - \frac{1}{2}} ⎤ ⎥ ⎥ ⎥ ⎦$
$V = 2 k q$

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An infinite number of charges, each numerically equal to q 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 due to this set of charges is ( Assume that 14πϵ0=k)

The correct option is A 2kqPotential at origin due to other charges V=kq1+kq2+kq4+kq8+..... V=kq[1+12+14+18+.....] V=kq⎡⎢ ⎢ ⎢⎣11−12⎤⎥ ⎥ ⎥⎦ V=2kq

An infinite number of charges, each numerically equal to $q$ 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 $x = 0$ due to this set of charges is ( Assume that $\frac{1}{4 π ϵ_{0}} = k$ )