SATHEE: electrostatic-potential-and-capacitance Question 21

electrostatic-potential-and-capacitance Question 21

Question: Q. 6. (i) Three point charges $q, - 4 q$ and $2 q$ are placed at the vertices of an equilateral triangle $ABC$ of side ’ $l$ ’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge $q$ .

(ii) Find out the amount of the work done to separate the charges at infinite distance.

A [Delhi/OD CBSE 2018]

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Solution:

Ans. (i) Finding the magnitude of the resultant force on charge $q$

(ii) Finding the work done

(i) Force on charge $q$ due to the charge $- 4 q$

$\vec{F_{1}} = \frac{1}{4 π ε_{0}} (\frac{4 q^{2}}{l^{2}}), along A B$

Force on the charge $q$ , due to the charge 2

$$ \vec{F}{2}=\frac{1}{4 \pi \varepsilon{0}}\left(\frac{2 q^{2}}{l^{2}}\right), \text { along } C A $$

The forces $F_{1}$ and $F_{2}$ are inclined to each other at an angle of $120^{\circ}$

Hence, resultant electric force on charge $q$

$\begin{aligned} F & = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ} & = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos 120^{\circ}} & = \sqrt{F_{1}^{2} + F_{2}^{2} - F_{1} F_{2}} & = (\frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l^{2}}) \sqrt{16 + 4 - 8} \end{aligned}$

$= \frac{1}{4 π ε_{0}} (\frac{2 \sqrt{3} q^{2}}{l^{2}})$

(ii) Net P.E. of the system

$\begin{aligned} = \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} [- 4 + 2 - 8] & = \frac{(- 10)}{4 π ε_{0}} \cdot \frac{q^{2}}{l} ∴ Work done & = \frac{10 q^{2}}{4 p ε_{0} l} = \frac{5 q^{2}}{2 π ε_{0} l} \end{aligned}$

[CBSE Marking Scheme 2018]

Detailed Answer :

(ii) The amount of work done to separate the charges at infinite distance is equal to the (-ve) potential energy of the given system. Now we know that the potential energy of three (03) charges at the corners of an equilateral triangle $A B C$ of side $l$ is given by

$U_{P E} = \frac{1}{4 π ε_{0}} [\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}}]$

given $q_{1} = q, q_{2} = - 4 q, q_{3} = 2 q, r_{12} = r_{13} = r_{23} = l$

$\begin{aligned} U_{P E} & = \frac{1}{4 π ε_{0}} [\frac{q (- 4 q)}{l} + \frac{q \cdot 2 q}{l} + \frac{2 q (- 4 q)}{l}] & = \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{l} [- 4 + 2 - 8] U_{P E} & = - \frac{10 q^{2}}{4 π ε_{0} l} \end{aligned}$

Therefore work done, $W = - U_{P E} = - (\frac{- 10 q^{2}}{4 π ε_{0} l})$

$\Rightarrow W = + \frac{5 q^{2}}{2 π ε_{0} l}$

Electrostatic Potential And Capacitance

electrostatic-potential-and-capacitance Question 21

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Electrostatic Potential And Capacitance

electrostatic-potential-and-capacitance Question 21

Detailed Answer :

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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