A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 ... upto infinity and a charge “-q” is fixed at each of the points x = 2x0, x = 4x0, x = 6x0 ... up to infinity. Here x0 is a positive constant. The potential at the origin of this system of charges is:

V=14πε0qx0+q3x0+q5x0+....upto infinity + 14πε0-qx0+q3x0+-q5x0+....upto infinity ⇒V=14πε0qx01-12+13-14+15-16....upto infinity Using, ln(1+x)=x-x22+x33-x44............. ⇒V =q4πε0x0ln(1+1)=q ln24πε0x0

A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 ... upto infinity and a charge “-q” is fixed at each of the points x = 2x0, x = 4x0, x = 6x0 ... up to infinity. Here x0 is a positive constant. The potential at the origin of this system of charges is:

V=14πε0qx0+q3x0+q5x0+....upto infinity + 14πε0-qx0+q3x0+-q5x0+....upto infinity ⇒V=14πε0qx01-12+13-14+15-16....upto infinity Using, ln(1+x)=x-x22+x33-x44............. ⇒V =q4πε0x0ln(1+1)=q ln24πε0x0

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A charge +q is fixed at each of the points x = x₀, x = 3x₀, x = 5x₀ ... upto infinity and a charge “-q” is fixed at each of the points x = 2x₀, x = 4x₀, x = 6x₀ ... up to infinity. Here x₀ is a positive constant. The potential at the origin of this system of charges is:

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zero

$\frac{q}{4 {πε}_{0} x_{0} \ln (2)}$

$\frac{qln (2)}{4 {πε}_{0} x_{0}}$

infinity

answer is D.

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Detailed Solution

$V = \frac{1}{4 {πε}_{0}} \{\frac{q}{x_{0}} + \frac{q}{3 x_{0}} + \frac{q}{5 x_{0}} + . . . . upto infinity\} + \frac{1}{4 {πε}_{0}} \{\frac{- q}{x_{0}} + \frac{q}{3 x_{0}} + \frac{- q}{5 x_{0}} + . . . . upto infinity\} \Rightarrow V = \frac{1}{4 {πε}_{0}} \frac{q}{x_{0}} \{1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} . . . . upto infinity\} [Using, \ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} . . . . . . . . . . . . .] \Rightarrow V = \frac{q}{4 {πε}_{0} x_{0}} \ln (1 + 1) = \frac{q \ln (2)}{4 {πε}_{0} x_{0}}$