SATHEE: electrostatic-potential-and-capacitance Question 20

electrostatic-potential-and-capacitance Question 20

Question: Q. 4. Obtain the expression for the potential due to an electric dipole of dipole moment $p$ at a point ’ $d$ ’ on the axial line.

A [O.D. Comptt. I, II, III 2013 ]

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Solution:

Ans. Consider an electric dipole of charges $+ q$ and $- q$ separated by $2 x$ distance being placed in free space. Let $P$ be the point at which the electric field is to be determined due to the electric dipole.

Let the potential at point $P$ due to positive charge be $V_{+}$ and the potential at point $P$ due to negative charge $V_{-}$

$\begin{aligned} V_{+} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{(d - x)} & V_{-} = \frac{1}{4 π ε_{0}} \cdot \frac{- q}{(d + x)} \end{aligned}$

$∴$ The total potential at point $P$ is given by

$\begin{aligned} V & = V_{+} + V_{-} V & = \frac{1}{4 π ε_{0}} (\frac{q}{d - x} - \frac{q}{d + x}) V & = \frac{1}{4 π ε_{0}} \cdot \frac{2 x q}{(d^{2} - x^{2})} ∴ 2 q x & = p V & = \frac{1}{4 π ε_{0}} \cdot \frac{p}{(d^{2} - x^{2})} \end{aligned}$

Now, if $x « < d$

$\begin{aligned} x^{2} & ≅ 0 V & = \frac{1}{4 π ε_{0}} \cdot \frac{p}{d^{2}} \end{aligned}$

AT Q. 5. Four point charges $Q, q, Q$ and $q$ are placed at the corners of a square of side ’ $a$ ’ as shown in the figure.

Find the

(i) resultant electric force on a charge $Q$ , and

(ii) potential energy of this system.

[Delhi/O.D. CBSE 2018]

Ans. (i) Finding the resultant force on a charge $Q 2$

(ii) Potential Energy of the system

(i) Let us find the force on the charge $Q$ at the point $C$ Force due to the other charge $Q$

$\begin{aligned} F_{1} = \frac{1 Q Q^{2}}{4 π ε_{0}} \cdot \frac{Q^{2}}{(a \sqrt{2})^{2}} & \vec{F_{1}} = \frac{1}{4 π ε_{0}} (\frac{Q^{2}}{2 a^{2}}) (along A C) \end{aligned}$

$1 / 2$

Force due to the charge $q$ (at $B$ ),

$$ \vec{F}{2}=\frac{1}{4 \pi \varepsilon{0}} \cdot \frac{q Q}{a^{2}} \text { along } B C $$

Force due to the charge $q$ (at $D$ ),

$\vec{F_{3}} = \frac{1}{4 π ε_{0}} \cdot \frac{q Q}{a^{2}} along D C$

Resultant of these two equal forces $\vec{F}{2} & \overrightarrow{F{3}}$

$\vec{F_{23}} = \frac{1}{4 π ε_{0}} \cdot \frac{q Q (\sqrt{2})}{a^{2}} (along A C)$

$∴$ Net force on charge $Q$ ( at point $C$ )

$$ \vec{F}=\vec{F}{1}+\overrightarrow{F{23}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{a^{2}}\left[\frac{Q}{2}+\sqrt{2} q\right] $$

$1 / 2$

This force is directed along $A C$ .

( For the charge $Q$ , at the point $A$ , the force will have the same magnitude but will be directed along $C A$ )

[Note : Don’t deduct marks if the student does not write the direction of the net force , $F]$

(ii) Potential energy of the system

$\begin{aligned} = \frac{1}{4 π ε_{0}} [4 \frac{q Q}{a} + \frac{q^{2}}{a \sqrt{2}} + \frac{Q^{2}}{a \sqrt{2}}] & = \frac{1}{4 π ε_{0} a} [4 q Q + \frac{q^{2}}{\sqrt{2}} + \frac{Q^{2}}{\sqrt{2}}] \end{aligned}$

[CBSE Marking Scheme 2018]

Electrostatic Potential And Capacitance

electrostatic-potential-and-capacitance Question 20

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Electrostatic Potential And Capacitance

electrostatic-potential-and-capacitance Question 20

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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