Electric Charges and Fields Test

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Electric Charges and Fields Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

The surface charge density on a copper sphere is $$\sigma$$. So the intensity of electric field on its surface will be:

A
$$\sigma$$

B
$$\dfrac{\sigma}{2}$$

C
$$\dfrac{\sigma}{2 \varepsilon o}$$

D
$$\dfrac{\sigma}{\varepsilon o}$$

Solution

The surface charge density on a copper sphere is $$\sigma $$
therefore$$,$$ the intensity of electric field on its surface will be$$-$$ $$\frac{\sigma }{{{ \in _0}}}$$
Hence,
option $$(D)$$ is correct answer.
Question 2
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At a certain distance from a point charge the electric field is $$500\ V/m$$ and the potential is $$3000\ V$$. What is this distance:

A
$$36\ m$$

B
$$12\ m$$

C
$$6\ m$$

D
$$144\ m$$

Solution

Given that,

Electric field $$E=500\,V/m$$

Potential $$V=3000V$$

We know that,

$$ E=\dfrac{V}{d} $$

$$ d=\dfrac{V}{E} $$

$$ d=\dfrac{3000}{500} $$

$$ d=6\,m $$

Hence, the distance is $$6\ m$$
Question 3
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An electron (of charge $$-e$$) revolves around a long wire with uniform charge density $$\lambda$$ in a circular path of radius $$r$$. Its kinetic energy is given by:

A
$$\dfrac { \lambda e }{ 8\pi { \epsilon }_{ 0 }r }$$

B
$$\dfrac { \lambda e }{ 4\pi { \epsilon }_{ 0 }r }$$

C
$$\dfrac { \lambda e }{ 2\pi { \epsilon }_{ 0 }r}$$

D
$$\dfrac { \lambda e }{ 4\pi { \epsilon }_{ 0 } }$$

Solution

Electric field for infinitely long wire is given by $$E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}r}\,where\,\,\lambda =ch\arg e\,\,density$$

Since the electron is revolving then the centripetal force is balanced by the electrostatic force

$$ {{F}_{e}}={{F}_{c}} $$

$$ eE=\dfrac{m{{v}^{2}}}{r} $$

$$ \dfrac{e\lambda }{2\pi {{\varepsilon }_{0}}}=m{{v}^{2}} $$

$$ \dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}}=\dfrac{1}{2}m{{v}^{2}} $$

$$ KE=\dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}} $$
Question 4
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Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is

A
$${3q/\epsilon_o}$$

B
$${2q/\epsilon_o}$$

C
$${q/\epsilon_o}$$

D
zero

Solution

Total net flux $$=0$$
But flux will be present in the region of surface $$S$$
Question 5
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The figure shows a family of equipotential surfaces and four paths along which an electron is made to move from one surface to another as shown in figure.
1) What is the direction of electric field?
2) Rank the paths according to the magnitude of the work done, greatest first

A
$$Rightward ; 4 > 3 > 2 > 1$$

B
$$Leftward ; 1 > 2 > 3 > 4$$

C
$$Rightward ; 3 = 4 > 2 = 1$$

D
$$Leftward ; 1 > 2 > 3 = 4$$

Solution

Along the direction of electric field, electric potential decreases, hence it should be from left to right.
Change in potential of path 3 and 4 is greater hence
$$ 3= 4 > 2 =1$$
Question 6
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Two charges are at a distance $$d$$ apart. If a copper plate of thickness $$d/2$$ is kept between them, then effective force will be

A
$$\dfrac{F}{2}$$

B
$$Zero$$

C
$$2F$$

D
$$\sqrt { 2F }$$

Solution

Copper plate is a conductor, so when a conductor is placed between the two charges then the force becomes zero because the value of dielectric constant is infinite. Force between the two charges in any medium is 1/K times the force between the same in air.
Question 7
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A point charge of value $$10^{-7}\ C$$ is situated at the centre of cube of $$1\ m$$ side. The electric flux through its total surface area is:

A
$$113\times 10^{4}\ Nm^{2}/C$$

B
$$11.3\times 10^{4}\ Nm^{2}/C$$

C
$$1.13\times 10^{4}\ Nm^{2}/C$$

D
$$none\ of\ these$$

Solution

Given that,

Charge $$q={{10}^{-7}}\,C$$

We know that,

The electric flux is

$$ \phi =\dfrac{q}{{{\varepsilon }_{0}}} $$

$$ \phi =\dfrac{{{10}^{-7}}}{8.85\times {{10}^{-12}}} $$

$$ \phi =0.1129\times {{10}^{5}} $$

$$ \phi =1.13\times {{10}^{4}}\,N{{m}^{2}}/C $$

Hence, the flux is $$1.13\times {{10}^{4}}\,N{{m}^{2}}/C$$
Question 8
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Position vectors of two point charges of $$1nC$$ each are $$(1,1,-1)$$m and $$(2,3,1)$$m respectively. Magnitude of electric force acting between them is ___N.

A
$${ 10 }^{ -3 }$$

B
$${ 10 }^{ -6 }$$

C
$${ 10 }^{ -9 }$$

D
$${ 10 }^{ -12 }$$

Solution

The Position vector of two point $$(1,1,-1)$$ and $$(2,3,1)$$

$$\vec r=(2-1)\hat i+(3-1)\hat j+(1+1)\hat k$$

$$\vec r=\hat i+2\hat j+2\hat k$$

$$r=|\vec r|=\sqrt{(1)^2+(2)^2+(2)^2}=3m$$

$$q_1=q_2=1nC=10^{-9}C$$

The magnitude of electric force is given by

$$F=\dfrac{1}{4\pi \varepsilon_0}.\dfrac{q_1 q_2}{r^2}$$

$$F=9\times 10^{9}\times \dfrac{10^{-9}\times 10^{-9}}{(3)^2}$$

$$F=9\times 10^{-9}\times \dfrac{1}{9}$$

$$F= 10^{-9}N$$
Question 9
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In the figure shown here, A is a conducting sphere and B is a closed spherical surface. If a-q change is placed at C near A, then the electric flux through the closed surface is -

A
zero

B
positive

C
negative

D
none of the above can be predicted

Solution

As the electric field inside the spherical conductor has to be zero the equal and opposite charges will be distributed on the conductor surface such that the induced electric field counters the electric field due to negative charge. For negative charges will be distributed on right and positive on left side
As the spherical region is on right side $$q$$ inside $$ < 0$$.
So, by Electric there $$=\dfrac {g\ inside}{E_0} < 0$$
Option $$C$$ is correct.
Question 10
1 / -0

A point charge of $$+6\mu C$$ is placed at a distance 20 cm directly above the centre of a square of side 40 cm. The magnitude of the flux through the square is

A
$$\epsilon_0$$

B
$$\dfrac{1}{\epsilon_0}$$

C
$$\epsilon_0\times 10^{-6}$$

D
$$\dfrac{1}{\epsilon_0} \times 10^{-6}$$

Solution

Given,
$$Q_{en}=+6\mu C=+6\times 10^{-6}C$$
From the Gauss's law, the magnetic flux through the cube is given by
$$\phi=\dfrac{Q_{en}}{\varepsilon_0}$$
The magnetic flux through the one face of the cube or square is,
$$\phi_s=\dfrac{Q_{en}}{6 \varepsilon_0}$$
$$\phi_s=\dfrac{6\times 10^{-6}}{6\varepsilon_0}$$
$$\phi_s=\dfrac{1}{\varepsilon_0}\times 10^{-6}$$
The correct option is D.