Electric Charges and Fields Test

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Electric Charges and Fields Test - 67

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Question 1
1 / -0

A charged particle 'q' lies at 'P' and the line PC is perpendicular to the surface of ABC (part of disc). Find the flux passing through the surface ABC.

A
$\displaystyle \frac { q }{ 4{ \varepsilon }_{ 0 } }$

B
$\displaystyle \frac { q }{ 16{ \varepsilon }_{ 0 } }$

C
$\displaystyle \frac { q }{ 32{ \varepsilon }_{ 0 } }$

D
$\displaystyle \frac { q }{ 48{ \varepsilon }_{ 0 } }$

Solution

$E=\cfrac{q}{4\pi \epsilon_0 a^2}$
Area of whole disk $=\pi r^2=\pi(\sqrt{3}a)^2=3\pi a^2$
Area of $ABC=\pi r^2\left( \cfrac { 30 }{ 360 } \right) =\cfrac { 3\pi { a }^{ 2 } }{ 12 }$
$=\cfrac { 1 }{ 12 }$ (area of whole disk)
The electric flux through the whole disk,
${ \phi }_{ T }=EA=\cfrac {q} {4\pi \epsilon_0 a^2}\times 3\pi a^2=\cfrac {3q}{4\epsilon_0}$
But electric flux, $\phi=\cfrac {1}{12}\phi_T=\cfrac{1}{12}\times\cfrac{3q}{4\epsilon_0}$
$=\cfrac {q}{16\epsilon_0}$
Question 2
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Given two positively charged particles $+Q$ & $+q$ . The $+Q$ charge is fixed in position, and the $+q$ charge is brought close to $+Q$ and released from rest. Choose the correct graph between acceleration $\left(a\right)$ of the $+q$ charge as a function of its distance $\left(r\right)$ from $+Q$ ?

A

B

C

D

E

Solution

The equation of motion of the charge q is $ma=F_{E}=\dfrac{kQq}{r^2}$ (By coulomb's law)
So, $a\propto \frac{1}{r^2}$
Thus, option A graph will correct representation.
Question 3
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If the charge on two-point charged bodies and the medium surrounding them are kept unchanged and the distance between them is reduced by $50$ % then the force between them _____.

A
is doubled

B
is quadrupled

C
becomes half

D
decreases to 1/4th of their original force

Solution

$F\propto \dfrac{1}{r^2}$

$R_1=R$
Since , distance is reduced to $50\%\implies R_2=\dfrac{R}{2}$

$\dfrac{F_2}{F_1}=\left(\dfrac{R_1}{R_2}\right)^2=4$

$\implies F_2=4F_1$

Hence force is quadrupled.

Answer-(B)
Question 4
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A charge $\displaystyle \left ( 5\sqrt{2}+2\sqrt{5} \right )$ coulomb is placed on the axis of an infinite disc at a distance a from the centre of disc. The flux of this charge on the part of the disc having inner and outer radius of a and 2a will be :

A
$\displaystyle \frac{3}{2\varepsilon _{0}}$

B
$\displaystyle \frac{1}{2\varepsilon _{0}}$

C
$\displaystyle \frac{2\left [ \sqrt{5}+\sqrt{2} \right ]}{\varepsilon _{0}}$

D
$\displaystyle \frac{2\sqrt{5}+5\sqrt{2}}{2\varepsilon _{0}}$

Solution

As solid angle, $w= 2\pi (1-cos\theta)$ where $\theta$ is half planer angle.
From trignometry, $tan \theta_1 = \dfrac{a}{a} = 1 \implies \theta_1 = 45^o$
Also $tan\theta_2 = \dfrac{2a}{a} = 2 \implies cos\theta_2 = \dfrac{1}{\sqrt{5}}$
Now solid angle subtended by shaded region, $w =w_2- w_1$
$w= 2\pi (1-cos\theta_2) - 2\pi (1-cos\theta_1)$ $= 2\pi (1-\dfrac{1}{\sqrt{5}}) - 2\pi (1-\dfrac{1}{\sqrt{2}})$
$w= 2\pi \bigg[\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{5}}\bigg]$
As flux through $4\pi$ steradian is equal to $\dfrac{q}{\epsilon_o}$
Thus flux through $w$ steradian, $\phi = \dfrac{q}{\epsilon_o} \times \dfrac{w}{4\pi}$
$\phi = \dfrac{2\sqrt{5} + 5\sqrt{2}}{\epsilon_o} \times \dfrac{2\pi [\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}]}{4\pi}$
$\phi = \dfrac{\sqrt{10} +5 -2 - \sqrt{10}}{2\epsilon_o}$
$\implies \phi = \dfrac{3}{2\epsilon_o}$
Question 5
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Two point charges $Q_1$ and $Q_2$ are positioned at points $1$ and $2$ , The field intensity to the right of the charge $Q_2$ on the line that passes through the two charges varies according to a law that is represented schematically in the figure.
The field intensity is assumed to be positive if its direction coincides with the positive direction on the $x-axis$ .The distance between the charges is $l$ . Find the sign of each charge.

A
$Q_2$ is positive and $Q_1$ is negative

B
$Q_2$ is positive and $Q_1$ is positive

C
$Q_2$ is negative and $Q_1$ is negative

D
$Q_2$ is negative and $Q_1$ is positive

Solution

$E=\dfrac{Q}{4\pi\epsilon_o r}$

Since, field intensity at point 2 that is at location of $Q_2$ is $-\infty$ .

Hence, $Q_2$ must be negative and only then for $r=0$ , field at 2 will be $-\infty$ .

Now at some distance right of point 2, field is 0.

At that point $E=\dfrac{Q_1}{4\pi\epsilon_o (l+a)}+\dfrac{Q_2}{4\pi\epsilon_o a}=0$

Since, $Q_2$ is negative and hence $Q_1$ must be positive for $E$ to be $0$ ..

Answer-(D)
Question 6
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Find the coulombic force between two $\alpha-particles$ separated by a distance of $8.2$ fermi, in air.

A
$10\ N$

B
$11\ N$

C
$14\ N$

D
$24\ N$

Solution

Each $\alpha-particle$ posses positive charge of $2e$

$e=1.6\times 10^{-19}C$
$Q=2e$ $R=8.2 fermi= 8.2\times 1 0^{-15}m$

$F=\dfrac{Q.Q}{4\pi\epsilon_o R^2}$

$\implies F=\dfrac{4e^2}{4\pi\epsilon_o R^2}$

$\implies F=\dfrac{9\times 10^9\times 4\times 1.6^2\times 10^{-38}}{8.2^2\times 10^{-30}}$

$\implies F=14N$

Answer-(C)
Question 7
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Mathematically, electric flux can $\phi$ be represented as:

$\vec E =$ electric field
$\hat n=$ surface normal vector
$A=$ surface area

A
$\vec \phi = E\hat nA$

B
$\vec \phi = \vec E.\hat nA$

C
$\phi =\vec E.\hat nA$

D
$\vec \phi = E nA$

Solution

$\phi=\vec{E}.\vec{A}$
Direction of area vector is taken along the normal to the surface.

Hence, $\phi=\vec{E}.\hat{n}A$ where $\hat{n}$ is direction normal to surface.

Answer-(C)
Question 8
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Calculate the flux through the base of the cone of radius $r$ .

A
$2\pi r^2 E$

B
$4\pi r^2 E$

C
$0.5\pi r^2 E$

D
$\pi r^2 E$

Solution

We know that electric flux is given bt

$\phi=\oint\vec{E}.\vec{dS}$
where $\vec{dS}$ is area vector directed along normal to area element.

For base of cone we can see that field is perpendicular to base surface and hence area vector is along field.

Hence, $\phi=\oint EdS=E\oint dS$

$\implies \phi=E(\pi r^2)$

$\implies \phi=\pi r^2E$

Answer-(D)
Question 9
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The electrostatic force of repulsion between two equal positively charged ions is $51.84\times 10^{-9}N$ , when they are separated by a distance of $0.4\ nm$ . How many electrons are missing from each ion?

A
$6$

B
$5$

C
$7$

D
$3$

Solution

$q_1=q_2=q(say)$

$r=0.4nm=4\times 10^{-10}m$

$F=51.84\times 10^{-9}N$

$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}$

$\implies 9\times 10^9\times \dfrac{q^2}{16\times 10^{-20}}=51.84\times 10^{-9}$

$\implies q^2=16\times 5.76 \times 10^{-38}$

$\implies q=4\times 2.4\times 10^{-19}C$

N=number of electrons

$q=Ne\implies N=\dfrac{q}{e}$

$N=\dfrac{4\times 2.4\times 10^{-19}}{1.6\times 10^{-19}}$

$\implies N=6$
Question 10
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Find the excess (equal in number) of electrons that must be placed on each of two small spheres spaced $3\ cm$ apart with a force of repulsion between the spheres to be $10^{-19}N$ .

A
$600$

B
$625$

C
$525$

D
$125$

Solution

$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}$

Given here that- $q_1=q_2=q(say)$
$r=3cm=3\times 10^{-2}m$

$\dfrac{1}{4\pi\epsilon_o}=9\times 10^9$
And, $F=10^{-19}N$

Hence $9\times 10^9\times \dfrac{q^2}{9\times 10^{-4}}=10^{-19}$

$\implies q^2\times 10^{13}=10^{=-19}$

$\implies q^2=10^{-32}$

$\implies q=10^{-16}C$

Let N=number of electrons

$q=Ne\implies N=\dfrac{q}{e}$

$\implies N=\dfrac{10^{-16}}{1.6\times 10^{-19}}$

$\implies N=625$

Answer-(B)