SATHEE: electric-charges-and-fields Question 23

electric-charges-and-fields Question 23

Question: Q. 6. Two point charges $+ q$ and $- 2 q$ are placed at the vertices ’ $B$ ’ and ’ $C$ ’ of an equilateral triangle $A B C$ of side ’ $a$ ’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex A due to these two charges.

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Solution:

Ans. (i) The magnitude

$\begin{aligned} | \vec{E_{A B}} | = \frac{1}{4 π ε_{0}} \cdot \frac{q}{a^{2}} = E & | \vec{E_{A C}} | = \frac{1}{4 π ε_{0}} \cdot \frac{2 q}{a^{2}} = 2 E \end{aligned}$

$\begin{aligned} E_{net} = \sqrt{4 E^{2} + E^{2} - 2 E^{2}} & E_{net} = E \sqrt{3} = \frac{1}{4 π ε_{0}} \frac{q \sqrt{3}}{a^{2}} \end{aligned}$

(ii) The direction of resultant electric field at vertex $A$

$\begin{aligned} \tan α & = \frac{E_{A B} \sin 120^{\circ}}{E_{A C} + E_{A B} \cos 120^{\circ}} \tan α & = \frac{E \times \frac{\sqrt{3}}{2}}{2 E + E \times (\frac{- 1}{2})} = \frac{1}{\sqrt{3}} α & = 30^{\circ} (with side A C) \end{aligned}$

[CBSE Marking Scheme, 2014]

[AT Q. 7. A charge is distributed uniformly over a ring of radius ’ $a$ ‘. Obtain an expression for the electric intensity $E$ at a point on the axis of the ring. Hence, show that for points at large distances from the ring, it behaves like a point charge.

A [Delhi I, II, III, 2016]

$O P = \sqrt{a^{2} + x^{2}}$ , by Pythagoras theorem

where, $a$ is radius of ring

$x$ is the distance from centre to point ’ $P$ '

$\begin{aligned} d E & = \frac{k d Q}{a^{2} + x^{2}} d E_{x} & = \frac{k d Q}{(a^{2} + x^{2})} \cos θ \end{aligned}$

and $d E_{y}$ will be zero because all components will cancel out each other.

$\begin{aligned} \cos θ & = \frac{x}{{(a^{2} + x^{2})}^{1 / 2}} d E_{x} & = \frac{k d Q}{a^{2} + x^{2}} \cdot \frac{x}{{(a^{2} + x^{2})}^{1 / 2}} E_{x} & = \int \frac{k x d Q}{{(a^{2} + x^{2})}^{3 / 2}} E_{x} & = \frac{k x}{{(a^{2} + x^{2})}^{3 / 2}} \int d Q E & = \frac{k Q x}{{(a^{2} + x^{2})}^{3 / 2}} \end{aligned}$

Now, if $x » a$

$E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{x^{2}}$ , which is equal to field produced by a point change.

$∴$ For large distance, ring behaves as a point charge. 1

[CBSE Marking Scheme, 2016]

Electric Charges And Fields

electric-charges-and-fields Question 23

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Electric Charges And Fields

electric-charges-and-fields Question 23

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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