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Electrostatic Potential and Capacitance

Potential due to a System of Charges

A charge +q is fixed at each of the points x=x_{0}, x=3 x_{0},

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Medium

Solving time: 3 mins

A charge $+ q$ is fixed at each of the points $x = x_{0}, x = 3 x_{0}$ , $x = 5 x_{0}$ ...upto infinite, on the $x$ -axis and a charge $- q$ is fixed at each of the points $x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0} \dots$ upto infinite. Here $x_{0}$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q / (4 π ε_{0} r)$ . Then, the potential at the origin due to the above system of charges is :

Name: Video solution 1: A charge +q is fixed at each of the points x=x0,x=3x0, x=5x0...upto infinite, on the x-axis and a charge −q is fixed at each of the points x=2x0,x=4x0,x=6x0… upto infinite. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4πε0r). Then, the potential at the origin due to the above system of charges is :
Uploaded: 2023-04-10T19:08:19.000Z
Duration: 14 min

\frac{q}{8 π ε _{0} x _{0} l n 2}

\infty

\frac{q l n 2}{4 π ε _{0} x _{0}}

Views: 5,770 students

Updated on: Apr 10, 2023

Book:

Physics Galaxy Electrostatics and Current Electricity Vol 3 A (Ashish Arora)

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Emily Discussed

A charge

+ q

is fixed at each of the points

x = x_{0}, x = 3 x_{0}

x = 5 x_{0}

...upto infinite, on the

x

-axis and a charge

- q

is fixed at each of the points

x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0} \dots

upto infinite. Here

x_{0}

is a positive constant. Take the electric potential at a point due to a charge

Q

at a distance

r

from it to be

Q / (4 π ε_{0} r)

. Then, the potential at the origin due to the above system of charges is :

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9 mins ago

Text solutionVerified

(D) Potential due to the given system of charges at origin is given as

V = \frac{1}{4 π ϵ _{0}} [\frac{q}{x _{0}} - \frac{q}{2 x _{0}} + \frac{q}{3 x _{0}} - \frac{q}{4 x _{0}} + \dots \infty] V = \frac{q}{4 π ϵ _{0} x _{0}} ln (1 + 1) V = \frac{q l n 2}{4 π ϵ _{0} x _{0}}

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Question Text	A charge $+ q$ is fixed at each of the points $x = x_{0}, x = 3 x_{0}$ , $x = 5 x_{0}$ ...upto infinite, on the $x$ -axis and a charge $- q$ is fixed at each of the points $x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0} \dots$ upto infinite. Here $x_{0}$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q / (4 π ε_{0} r)$ . Then, the potential at the origin due to the above system of charges is :
Updated On	Apr 10, 2023
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	301
Avg. Video Duration	10 min

Text solutionVerified

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Text solutionVerified

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