physics-class-12-unit-02-chapter-07-problem-solving-electrostatics-l-7-9-qsbqljhthga

<h3 id="success-stylefont-size25pxproblem-solving-electrostatics--l-7-success-1"><Success style="font-size:25px;">Problem-Solving-Electrostatics  L-7 </Success></h3>
<h3 id="primary-stylefont-size24pxproblem-1primary"><primary style="font-size:24px;">Problem-1</primary></h3>
<hr>
<main>
<div style='float: left; width:60%;font-size:28px;text-align:left;'>
<ul>
<li>
<p>A Wooden block performs SHM on a frictionless surface with frequency $v_0$. The block contains a charge +q on its surface. If now, a uniform electric field $\vec{E}$ is switched on as shown, then SHM of the block will be (2011)</p>
</li>
<li>
<p>A. of the same frequency and with shifted mean position.</p>
</li>
<li>
<p>B. of the same frequency and with the same mean position.</p>
</li>
<li>
<p>C. of changed frequency and with shifted mean position.</p>
</li>
<li>
<p>D. of changed frequency with same mean position.</p>
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</ul>
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<footer style = "font-size: 18px">  $\rightarrow$ Problems on Electrostatics $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution-1 $\rightarrow$ Solution </footer>

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<h3 id="primary-stylefont-size24pxsolution-1primary"><primary style="font-size:24px;">Solution-1</primary></h3>
<hr>
<main>
<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>Before $\vec{E}$ is switched on</p>
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<li>
<p>$\nu_0=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$</p>
</li>
<li>
<p>Now $\vec{E}$ is switched on:</p>
</li>
<li>
<p>Block is in equilibrium at 0</p>
</li>
<li>
<p>$q E=k x_0 \Rightarrow x_0=\frac{q E}{k}$</p>
</li>
</ul>
</div>
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<footer style = "font-size: 18px"> Problems on Electrostatics $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

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<hr>
<main>
<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>$m \frac{d^2 x}{d t^2}=-k(x+x_0)+q E$</p>
</li>
<li>
<p>$=-kx-kx_0+kx_0=-kx$</p>
</li>
<li>
<p>$\Rightarrow  \frac{d^2 x}{d t^2}+\frac{k}{m} x=0$</p>
</li>
<li>
<p>$ \omega=\sqrt{\frac{k}{m}} $</p>
</li>
<li>
<p>$\gamma =\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Problem-1 $\rightarrow$ Solution-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution-2 </footer>

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<h3 id="primary-stylefont-size24pxproblem-2primary"><primary style="font-size:24px;">Problem-2</primary></h3>
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<main>
<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>Consider a system of three charges $\frac{q}{3},\frac{q}{3}and-\frac{2q}{3}$ placed at point A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB=$60^o$.(2006)</li>
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<footer style = "font-size: 18px"> Solution-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Problem $\rightarrow$ Solution-2 </footer>

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<hr>
<main>
<div style='float: left; width:100%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>$q_A =q / 3, q_B=q/3, q_C=\frac{-2 q}{3}$</p>
</li>
<li>
<p>$\vec{E}_0 =-\frac{2 q / 3}{4 \pi \epsilon_0 R^2} \hat{x}=-\frac{q}{6 \pi \epsilon_0 R^2}\hat{x}$</p>
</li>
<li>
<p>$U=\frac{1}{4 \pi \epsilon_0}[\frac{q_A q_B}{r_{A B}}+\frac{q_A q_C}{r_{A C}}+\frac{q_B q_C}{r_{B C}}]$</p>
</li>
<li>
<p>$ r_{A B} = 2 R, r_{A C} =r_{A B} \sin 30^o$</p>
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<li>
<p>$=\frac{1}{2} r_{A B}=R$</p>
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<li>
<p>$r_{BC}=\sqrt{3}R$</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Problem-2 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution-2 </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

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<h3 id="primary-stylefont-size24pxproblem-3primary"><primary style="font-size:24px;">Problem-3</primary></h3>
<hr>
<main>
<div style='float: left; width:60%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$, as shown in the figure. Given that $K=\frac{1}{4 x \epsilon_0} \frac{q}{L^2}$, which of the following statement(s) is (are) correct?</p>
</li>
<li>
<p>A. The electric field at O is 6K along OD.</p>
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<li>
<p>B. The potential at O is zero.</p>
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<li>
<p>C. The potential at all points on the line PR is same.</p>
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<li>
<p>D. The potential at all points on the line ST is same.</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Solution-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Solution-3 </footer>

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<hr>
<main>
<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>The elctric field at o</p>
</li>
<li>
<p>$E_{OA} = \frac{2q}{4 \pi \epsilon_0 L^2} = 2K$</p>
</li>
<li>
<p>Due to chargesat A and D is 4K along OD</p>
</li>
<li>
<p>Due to charges at B and E is 2K along OE</p>
</li>
<li>
<p>Due to charges at C and F is 2K along OC</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Problem-4 </footer>

<h3 id="success-stylefont-size25pxproblem-solving-electrostatics--l-7-success-10"><Success style="font-size:25px;">Problem-Solving-Electrostatics  L-7 </Success></h3>
<h3 id="primary-stylefont-size24pxsolution-3primary-1"><primary style="font-size:24px;">Solution-3</primary></h3>
<hr>
<main>
<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>
<p><strong>Resultant force</strong></p>
</li>
<li>
<p>$=2k \cos 60^o + 2K \cos 60^o +4K=6K$ along OD</p>
</li>
<li>
<p><strong>Potential at O</strong></p>
</li>
<li>
<p>$V_0= \sum \frac{1}{4 \pi \epsilon_0} \frac{qi}{L}=\frac{1}{4 \pi \epsilon_0 L} \sum q_i$=0</p>
</li>
<li>
<p><strong>(A) (B) (C) correct option</strong></p>
</li>
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<footer style = "font-size: 18px"> Problem-3 $\rightarrow$ Solution-3 $\rightarrow$ <span style = "color:blue"> Solution-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution-4</footer>

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<h3 id="primary-stylefont-size24pxproblem-4primary"><primary style="font-size:24px;">Problem-4</primary></h3>
<hr>
<main>
<div style='float: left; width:100%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>Under the influence of the Coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement(s).(2009)</p>
</li>
<li>
<p>A. The angular momentum of the charge $-q$ is constant.</p>
</li>
<li>
<p>B. The linear momentum of the charge $-q$ is constant.</p>
</li>
<li>
<p>C. The angular velocity of the charge $-q$ is constant.</p>
</li>
<li>
<p>D. The linear speed of the charge $-q$ is constant.</p>
</li>
</ul>
</div>
<div style='float: right; width:0%;'>


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<hr>
<footer style = "font-size: 18px"> Solution-3 $\rightarrow$ Solution-3 $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution-4 $\rightarrow$ Problem-5 </footer>

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<hr>
<main>
<div style='float: left; width:60%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>Torque on charge $-q$ due to coulomb force</p>
</li>
<li>
<p>$ \vec{F}=\frac{-Q q}{4 \pi \epsilon_0 r^2} \hat{r} $</p>
</li>
<li>
<p>$ \vec{\tau}=\vec{r} \times \vec{F}=0 $</p>
</li>
<li>
<p>$ \Rightarrow \frac{d \vec{L}}{d t}=0  \Rightarrow \vec{L} \equiv $  constant</p>
</li>
<li>
<p>$ \vec{F}=\frac{d \vec{p}}{d t} \neq 0$</p>
</li>
<li>
<p>$\Rightarrow \vec{P} $  is not constant</p>
</li>
<li>
<p>$ L  =m \omega r^2$</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Solution-3 $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution-4 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution-5 </footer>

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<h3 id="primary-stylefont-size24pxproblem-5primary"><primary style="font-size:24px;">Problem-5</primary></h3>
<hr>
<main>
<div style='float: left; width:100%;font-size:30px;text-align:left;'>
<ul>
<li>Four point charges, each of $+q$, are rigidly fixed at the four corners of a square planar soap film of side a. The surface tension of soap film is $\gamma$. The system of charges and planar film are in equilibrium and $a=k[q^2 / \gamma]^{1 / N}=$ where $k$ is a constant. Then $N$ is …….
(2011)</li>
</ul>
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<footer style = "font-size: 18px"> Problem-4 $\rightarrow$ Solution-4 $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution-5 $\rightarrow$ Solution </footer>

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<hr>
<main>
<div style='float: left; width:100%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>The force on line BC due to surface tension is $\gamma a$</p>
</li>
<li>
<p>$K=\frac{1}{4 \pi \epsilon_0}$</p>
</li>
<li>
<p>$ F_1=\frac{q^2}{4 \pi \epsilon_0 a^2}=\frac{K q^2}{a^2}=F_2 $</p>
</li>
<li>
<p>$ F_3=\frac{q^2}{4 \pi \epsilon_0 2 a^2}=\frac{K q^2}{2 a^2}$</p>
</li>
<li>
<p><strong>Resultant force at A:</strong></p>
</li>
<li>
<p>$F=2 \frac{k q^2}{a^2} \cos 45^o +\frac{k q^2}{2 a^2}$</p>
</li>
<li>
<p>$F=\frac{k-\epsilon^2}{a^2}[\sqrt{2}+\frac{1}{2}]$</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Solution-4 $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution-5 </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

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<div style='float: left; width:100%;font-size:30px;text-align:left;'>
<ul>
<li>
<p>$\gamma a = 2F \cos 45^o$</p>
</li>
<li>
<p>$\gamma a  =\sqrt{2} \frac{k q^2}{a^2}[\sqrt{2}+\frac{1}{2}]$</p>
</li>
<li>
<p>$ \Rightarrow a^3 =\sqrt{2}K(\sqrt{2}+\frac{1}{2})\frac{q^2}{\gamma} = K_0 \frac{q^2}{\gamma}$</p>
</li>
<li>
<p>$\Rightarrow a = k (\frac{q^2}{\gamma})^{1/3}$</p>
</li>
<li>
<p>$\Rightarrow a=k(\frac{q^2}{\gamma})^{1 / 3}$</p>
</li>
<li>
<p>$N=3$</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Problem-5 $\rightarrow$ Solution-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Problem </footer>

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<div style='float: left; width:50%;font-size:30px;text-align:left;'>
<ul>
<li>
<p><strong>Matrix or Matching Type</strong></p>
</li>
<li>
<p>Four charges $Q_1,Q_2,Q_3$ and $Q_4$ of same magnitude are fixed along the x axis at x = -2a, -a, a and +2a, respectively. A positive charge q is placed on the positive y axis at a distance b>0. Four options of the sign of these charges are given in Colomn I. The direction of the forces on the charge q is the charge q is given is Column II. Match Column I with Column II.(2014)</p>
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<footer style = "font-size: 18px"> Solution-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem $\rightarrow$ Solution-6 </footer>

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<hr>
<main>
<div style='float: left; width:100%; font-size:20px; text-align:left;'>
<table>
<thead>
<tr>
<th>Column 1</th>
<th>Column 2</th>
</tr>
</thead>
<tbody>
<tr>
<td>(P) $q_1,q_2,q_3,q_4$ all positive</td>
<td>+z</td>
</tr>
<tr>
<td>(Q) $q_1,q_2$ positive, $q_3,q_4$ negative</td>
<td>-z</td>
</tr>
<tr>
<td>(R) $q_1,q_4$ positive, $q_2,q_3$ negative</td>
<td>+y</td>
</tr>
<tr>
<td>(S) $q_1,q_3$ positive, $q_2,q_4$ negative</td>
<td>-y</td>
</tr>
</tbody>
</table>
<ul>
<li>A. $P-3, Q-1, R-4, S-2$</li>
<li>B. $P-4, Q-2, R-3, S-1$</li>
<li>C. $P-3, Q-1, R-2, S-4$</li>
<li>D. $P-4, Q-2, R-1, S-3$</li>
</ul>
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<footer style = "font-size: 18px"> Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution-6 $\rightarrow$ Solution-6 </footer>

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<div style='float: left; width:50%;font-size:25px;text-align:left;'>
<ul>
<li>$|q_1|=|q_2|=|q_3|=|q_4|=q`$</li>
</ul>
<ul>
<li>
<p>Resulting force is along $+\hat{y}$ direction</p>
</li>
<li>
<p>Resultant force is along $+\hat{x}$ direction</p>
</li>
</ul>
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<footer style = "font-size: 18px"> Problem-6 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue">Solution-6 </span> $\rightarrow$ Solution-6 $\rightarrow$ Problem-7 </footer>

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<p><strong>Integer Type</strong></p>
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<p>A particle of mass $10^{-1}$ kg and charge 1.0C, is initially at rest. At time t=0, the particle comes under the influence of an electric field $\vec{E}(t)=E_0 \sin \omega ti $, where $E_0 = 1.0N/C$ and $\omega = 10^3 rad/s$. Consider the effect of only the elctric force on the particle. Then the maximum speed, in m/s, attained by the particle at subsequent times is …. (201.)</p>
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<footer style = "font-size: 18px"> Solution-6 $\rightarrow$ Solution-6 $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Solution-7 $\rightarrow$ Problem-8 </footer>

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<h3 id="primary-stylefont-size24pxsolution-7primary"><primary style="font-size:24px;">Solution-7</primary></h3>
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<p>$\vec{F}=q \vec{E}$</p>
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<p>$ \vec{a}=\frac{\vec{F}}{m}=\frac{q \vec{E}}{m}$</p>
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<p>$ \rightarrow \frac{d \vec{v}}{d t}=\frac{q E_0 \sin \omega t}{m} \hat{i}$</p>
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<p>$\rightarrow \int_{\vec {V}_0}^{\vec{v}} d \vec{v}=\int_0^t \frac{q E_0}{m} \sin \omega t dt \hat{i} = \frac{q E_0}{m} \hat{i} \int_0^t \sin \omega t d t$</p>
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<p>$\Rightarrow \vec{v} - \vec{v}_0=\frac{q E_0}{m} \hat{i} [-\frac{\cos \omega t}{\omega}]|_0^t$</p>
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<p>$\Rightarrow \vec{v}=-\frac{q E_0}{m \omega} \hat{i}[\cos \omega t-1]$</p>
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<p>$\Rightarrow \vec{v}=\hat{i} \frac{q E_0}{m \omega}(1-\cos \omega t)=\hat{i} \frac{2 q E_0}{m \omega}$</p>
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<p>$max^m$ speed $ V_{max} =\frac{2 q E_0}{m \omega}$ $ =2 m/s $</p>
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<footer style = "font-size: 18px"> Solution-6 $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution-7 </span> $\rightarrow$ Problem-8 $\rightarrow$ Solution-8 </footer>

<h3 id="success-stylefont-size25pxproblem-solving-electrostatics--l-7-success-22"><Success style="font-size:25px;">Problem-Solving-Electrostatics  L-7 </Success></h3>
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<p>Charges $q$, $2_q$ and $4_q$ are uniformaly distributed in three electric solid sphere 1, 2, ans 3 of radii $i/2$, r and 2r respectively, as shown in the figure. If magnitude of the electric fields ojn point P as a distance r from the centre of sphere 1, 2, and 3 are $E_1$, $E_2$ and $E_3$ respectively them (2014)</p>
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<p>A. $E_1>E_2>E_3$</p>
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<p>B. $E_3>E_1>E_2$</p>
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<p>C. $E_2>E_1>E_3$</p>
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<p>D. $E_3>E_2>E_1$</p>
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<p>$ E_1 \times 4 \pi r^2=\frac{q}{\epsilon_0} $</p>
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<p>Sphere 1 $ \Rightarrow E_1=\frac{q}{4 \pi \epsilon_0 r^2}$</p>
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<p>Sphere 2 $ E_2 \times 4 \pi r^2=\frac{2 q}{\epsilon_0}$</p>
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<p>$\Rightarrow  E_2=2 \frac{2q}{4 \pi \epsilon_0 r^2}=2 E_1$</p>
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<p>sphere 3 $q_{enc}=\frac{4q}{4/3 \pi (2r)^3}\times \frac{4}{3}\pi r^3$</p>
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<p>$=\frac{q}{2}$</p>
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<p>$E_3 \times 4 \pi r^2=\frac{q_{enc}}{\epsilon_0} \rightarrow E_3 =\frac{1}{2} \frac{q}{4 \pi \epsilon_0 r^2}$</p>
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<p>$=\frac{1}{2}E_1$</p>
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<p><strong>(C) correct option</strong></p>
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<main>
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<p>Consider an electirc field $\vec{E}=E_0\hat{i}$ is a constant. The flux through the shaded region (as shown in the figure) due to the field is (2011)</p>
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<p>$(A) 2E_0a^2$</p>
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<p>$(B)\sqrt{2}E_0a^2$</p>
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<p>$(C)E_0a^2$</p>
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<p>$(D)\frac{E_0a^2}{2}$</p>
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<div style='float: left; width:100%;font-size:30px;text-align:left;'>
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<p>$\oint=\oint \vec{E}.\vec{ds}=\vec{E}. \oint \vec{ds} = \vec{E}.\vec{S}$</p>
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<p>$\vec{S}=(a \hat{j}) \times (a \hat{i} +a \hat{k}) = a^2(-\hat{k}+\hat{i})=a^2(\hat{i}-\hat{k})$</p>
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<p>$\oint =E_0 \hat{i}.a^2(\hat{i}-\hat{k})=E_0 a^2$</p>
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