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
1 / -0

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)