Electric Charges and Fields Test

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Electric Charges and Fields Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Two equal point charges each of $$3\mu C$$ are separated by a certain distance in metres. If they are located at $$\left( i+j+k \right) $$ and $$\left( 2i+3j+k \right) $$, then the electrostatic force between them is

A
$$16.2 \times { 10 }^{ -3 }N$$

B
$$9\times { 10 }^{ -3 }N$$

C
$${ 10 }^{ -3 }N$$

D
$$9\times { 10 }^{ -2 }N$$

E
$$3\times { 10 }^{ -3 }N$$

Solution

The distance between two charges is $$r=\sqrt{(2-1)^2+(3-1)^2+(1-1)^2}=\sqrt 5$$
By Coulomb's law, the force between two charges is $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}=(9\times 10^9)(\dfrac{3\times 10^{-6}\times 3\times 10^{-6}}{5})=16.2\times 10^{-3} N$$
Question 2
1 / -0

A cylinder of radius r and length l is placed in a uniform electric field of intensity E acting parallel to the axis of the cylinder. The total flux over curved surface area is:

A
$$2\pi rE$$

B
$$\left(\displaystyle\frac{2\pi}{l}\right)E$$

C
$$2\pi rlE$$

D
$$\displaystyle\frac{E}{2\pi rl}$$

E
zero

Solution

Total flux emerging from the curved surface of the cylinder,
$$\phi =E\cdot ds=Eds\cos\theta$$
Here, $$\theta =90^o$$
$$\phi =Eds\cos 90^o$$
$$\phi =0$$.
Question 3
1 / -0

An electron is moving round the nucleus of hydrogen atom in a circular orbit of radius. The coulomb force F between the two is?

A
$$k\displaystyle\frac{e^2}{r^3} \vec r$$

B
$$-\displaystyle k\frac{e^2}{r^3} \vec r$$

C
$$\displaystyle k\frac{e^2}{r^2} \vec r$$

D
$$-\displaystyle k\frac{e^2}{Ar}$$

Solution

Using Coulomb's law, the force between two charges is given by,
$$\vec F=k\dfrac{q_1q_2}{r^3}\vec r$$ where $$r=$$ distance between charges.
Here, $$q_1=-e$$, charge of an electron ; $$q_2=Ze=e$$, charge of the nucleus and $$r=$$ radius of hydrogen atom.
(note: for a hydrogen atom, atomic number $$=Z=1$$)
Thus, the force between electron and nucleus will be, $$F=k\dfrac{(-e)e}{r^3}\vec r=-k\dfrac{e^2}{r^3}\vec r$$
Question 4
1 / -0

Electric field produced due to an infinitely long straight uniformly charged wire at perpendicular distance of $$2 cm$$ is $$3\times { 10 }^{ 8 }N{ C }^{ -1 }$$. Then, linear charge density on the wire is _____________.
$$\left( K=9\times { 10 }^{ 9 }SI\quad unit \right) $$

A
$$3.33\dfrac { \mu C }{ m } $$

B
$$333\dfrac { \mu C }{ m } $$

C
$$666\dfrac { \mu C }{ m } $$

D
$$6.66\dfrac { \mu C }{ m } $$

Solution

Given, $$r=2\times { 10 }^{ -2 }m$$,
$$E=3\times { 10 }^{ 8 }{ N }/{ C }$$
$$K=9\times { 10 }^{ 9 }{ N{ m }^{ 2 } }/{ { C }^{ 2 } }$$
The linear charge density $$I=\dfrac { E\cdot r }{ 2K } $$
$$\Rightarrow I=\dfrac { 3\times { 10 }^{ 8 }\times 2\times { 10 }^{ -2 } }{ 2\times 9\times { 10 }^{ 9 } } $$
$$\Rightarrow I=\dfrac { 1 }{ 3 } \times { 10 }^{ -3 }{ C }/{ m }$$
$$\Rightarrow =0.333\times { 10 }^{ -3 }{ C }/{ m }$$
$$=333\times { 10 }^{ -6 }{ C }/{ m }$$
$$=333{ \mu C }/{ m }$$
Question 5
1 / -0

The electrostatic force between two point charges is directly proportional to the

A
Sum of the charges

B
Distance between the charges

C
Permittivity of the medium

D
Square of the distance between the charges

E
Product of the charges

Solution

By Coulomb's law
$$F=\frac {1}{4\pi\varepsilon _0}\cdot \frac {q_1q_2}{r^2}$$
So, the electrostatic force between two point charges is directly proportional to the product of the charge.
Question 6
1 / -0

Two lithium nuclei in lithium vapour at room temperature do not combine to form a carbon nucleus because

A
Carbon nucleus is an unstable particle

B
It is not energetically favourable particle

C
Nuclei do not come very close due to Coulombic repulsion

D
Lithium nucleus is more tightly bound than a carbon nucleus

Solution

Due to Coulombic repulsion (i.e. nuclei are having same charge) these do not combine at room temperature.
Question 7
1 / -0

If two charges $$+4e$$ and $$+e$$ are at a distance $$x$$ apart then at what distance charge $$q$$ must be placed from $$+e$$, so that it is in equilibrium?

A
$$\dfrac {x}{2}$$

B
$$\dfrac {x}{3}$$

C
$$\dfrac {x}{6}$$

D
$$\dfrac {2x}{3}$$

Solution

For equilibrium of $$q$$
$$|F_{1}| = |F_{2}|$$
$$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {qq_{1}}{x_{1}^{2}} = \dfrac {1}{4\pi \epsilon_{0}} = \dfrac {qq_{2}}{x_{2}^{2}}$$
$$x_{2} = \dfrac {x}{\sqrt {\dfrac {q_{1}}{q_{2}}} + 1} = \dfrac {x}{\sqrt {\dfrac {4e}{e}} + 1} = \dfrac {x}{3}$$.
Question 8
1 / -0

There exists an electric field of $$1$$N$$/$$C along Y direction. The flux passing through the square of $$1$$m placed in XY plane inside the electric field is?

A
$$1.0Nm^2/C$$

B
$$10.0Nm^2/C$$

C
$$2.0Nm^2/C$$

D
Zero

Solution

The flux passing through the square of $$1$$m placed in XY plane inside the electric field is zero because by Gauss theorem we can say here closed circuit not formed.
Question 9
1 / -0

The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius $$10cm$$ surrounding the total charge is $$20Vm$$. The flux over a concentric sphere of radius $$20cm$$ will be

A
$$20V.m$$

B
$$25V.m$$

C
$$40V.m$$

D
$$200V.m$$

Solution

According to Gauss's law total flux coming out of a closed surface enclosing charge q is given by
ϕ=∮E .dS =qε0
From this expression, it is clear the total flux linked with a closed surface only depends on the enclosed charge and independent of the shape and size of the surface.
ϕ=∮E .dS =qε0=20Vm
this qε0 is constant as long as the enclosed charge is constant
The flux over a concentric sphere of radius 20cm=20Vm
Question 10
1 / -0

If a mass of $$20\ g$$ having charge $$3.0\ mC$$ moving with velocity $$20\ ms^{-1}$$ enters a region of electric field of $$80\ NC^{-1}$$ in the same direction as the velocity of mass, then the velocity of mass after $$3s$$ in the region will be

A
$$40\ ms^{-1}$$

B
$$44\ ms^{-1}$$

C
$$56\ ms^{-1}$$

D
$$80\ ms^{-1}$$

Solution

As the mass is moving in the electric field, then,
$$ma = qE$$
$$a = \dfrac {qE}{m} = \dfrac {3\times 10^{-3} \times 80}{20\times 10^{-3}} = 12ms^{-2}$$
By using, $$v = u + at$$
$$v = 20 + 12\times 3 = 56\ ms^{-1}$$.

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Electric Charges and Fields Test - 47

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Electric Charges and Fields Test - 47

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Question 1

$$16.2 \times { 10 }^{ -3 }N$$

$$9\times { 10 }^{ -3 }N$$

$${ 10 }^{ -3 }N$$

$$9\times { 10 }^{ -2 }N$$

$$3\times { 10 }^{ -3 }N$$

Solution

Question 2

$$2\pi rE$$

$$\left(\displaystyle\frac{2\pi}{l}\right)E$$

$$2\pi rlE$$

$$\displaystyle\frac{E}{2\pi rl}$$

zero

Solution

Question 3

$$k\displaystyle\frac{e^2}{r^3} \vec r$$

$$-\displaystyle k\frac{e^2}{r^3} \vec r$$

$$\displaystyle k\frac{e^2}{r^2} \vec r$$

$$-\displaystyle k\frac{e^2}{Ar}$$

Solution

Question 4

$$3.33\dfrac { \mu C }{ m } $$

$$333\dfrac { \mu C }{ m } $$

$$666\dfrac { \mu C }{ m } $$

$$6.66\dfrac { \mu C }{ m } $$

Solution

Question 5

Sum of the charges

Distance between the charges

Permittivity of the medium

Square of the distance between the charges

Product of the charges

Solution

Question 6

Carbon nucleus is an unstable particle

It is not energetically favourable particle

Nuclei do not come very close due to Coulombic repulsion

Lithium nucleus is more tightly bound than a carbon nucleus

Solution

Question 7

$$\dfrac {x}{2}$$

$$\dfrac {x}{3}$$

$$\dfrac {x}{6}$$

$$\dfrac {2x}{3}$$

Solution

Question 8

$$1.0Nm^2/C$$

$$10.0Nm^2/C$$

$$2.0Nm^2/C$$

Zero

Solution

Question 9

$$20V.m$$

$$25V.m$$

$$40V.m$$

$$200V.m$$

Solution

Question 10

$$40\ ms^{-1}$$

$$44\ ms^{-1}$$

$$56\ ms^{-1}$$

$$80\ ms^{-1}$$

Solution

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