A charge +q is fixed at each of the points x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.For th

Vedclass · Accepted Answer

$\frac{q \log_e 2}{4\pi\varepsilon_0 x_0}$

Vedclass · Answer

(D) The potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.For the given system,the charges $+q$ are at $x = x_0, 3x_0, 5x_0, \dots$ and charges $-q$ are at $x = 2x_0, 4x_0, 6x_0, \dots$.The potential at the origin is:$V = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{x_0} - \frac{1}{2x_0} + \frac{1}{3x_0} - \frac{1}{4x_0} + \dots \right)$$V = \frac{q}{4\pi\varepsilon_0 x_0} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right)$Using the Taylor series expansion for $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,for $x=1$,we get $\log_e(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.Substituting this into the expression for $V$:$V = \frac{q}{4\pi\varepsilon_0 x_0} \log_e 2$.

Similar Questions

Two identical positive charges are placed on the $y$-axis at $y=-a$ and $y=+a$. The variation of $V$ (electric potential) along the $x$-axis is shown by which graph?

Two tiny spheres carrying charges $1.8 \mu C$ and $2.8 \mu C$ are located $40 \ cm$ apart. The potential at the mid-point of the line joining the two charges is:

The minimum work needed to be done to bring a charge $q=6 \mu C$ from $\infty$ to a point $0.75 \ m$ from a charge $Q=30 \mu C$ is (in $J$)

Is electrostatic potential a vector or a scalar quantity?

$A$ charge of $10 \,\mu C$ is placed at the origin of an $x-y$ coordinate system. The potential difference between two points $(0, a)$ and $(a, 0)$ in volts will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module