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

$A$ regular hexagon of side $9 \ cm$ has a charge $5 \mu C$ at each of its vertices. The electric potential at its centre is $..........V$. $(k = 9 \times 10^9 \ Nm^2 C^{-2})$

$A$ hollow charged metal sphere has radius $R$. If the potential difference between its surface and a point at a distance $5R$ from the centre is $V$,then the magnitude of the electric field intensity at a distance $5R$ from the centre of the sphere is:

Two particles $A$ and $B$ having the same mass have charges $+q$ and $+4q$ respectively. When they are allowed to fall from rest through the same electric potential difference,the ratio of their speeds $v_A$ to $v_B$ will be:

For a point charge,the electric potential and electric field at a given point are $600 \, V$ and $200 \, N/C$ respectively. The magnitude of the point charge is ......... $\mu C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module