Electric Charges and Fields Test

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Electric Charges and Fields Test - 35

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Question 1
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Two pith balls each of mass $$1g$$ and carrying a charge 1$$\mu$$C are attached to the ends of silk threads $$1m$$ long, the other ends of which are attached to some fixed point, in a gravity free space. The force between them is :

A
$$9.8 \times 10^{-2} N$$

B
$$9.0 \times 10^{-3} N$$

C
$$4.5 \times 10^{-2} N$$

D
$$0.9 \times 10^{-3} N$$

Solution

As they are in a gravity free space so the force between them is only the electrostatic force. i.e,

$$F=\dfrac{q^2}{4\pi\epsilon_0r^2}=\dfrac{9 \times 10^9 (10^{-6})^2}{1^2}=9\times 10^{-3} N$$
Question 2
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The electric field in a region is given by $$\overrightarrow{E} = \dfrac{3}{5} E_0 \hat{j}$$ with $$E_0=2 \times 10^3 NC^{-1}$$ Find the flux of this field through a rectangular surface of area $$0.2 m^2$$ parallel to the $$Y-Z$$ plane.

A
$$320 Nm^2 C^{-1}$$

B
$$240 Nm^2 C^{-1}$$

C
$$400 Nm^2 C^{-1}$$

D
none of these

Solution

The electric flux , $$\phi=\vec E. \hat{n}dS$$
Since the surface is parallel to $$Y-Z$$ plane so the normal unit vector will be along x-axis i.e $$\hat n=\hat i$$
Thus, $$\phi=(3/5)E_0 \hat j. \hat i (0.2)=0$$ (as $$\hat j.\hat i=0$$)
Thus, option D will correct.
Question 3
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The electric field at a point $$5 cm$$ from a long line charge of density $$2.5 \times 10^{-6} cm^{-1}$$ is :

A
$$9 \times 10^3 NC^{-1}$$

B
$$9 \times 10^4 NC^{-1}$$

C
$$9 \times 10^5 NC^{-1}$$

D
$$9 \times 10^6 NC^{-1}$$

Solution

The electric field at a distance r from the long wire is

$$\displaystyle E=\dfrac{1}{4\pi \epsilon_0} \dfrac{2\lambda}{r}=9\times 10^9\times \dfrac{2\times 2.5\times 10^{-6}}{5\times 10^{-2}}=9\times 10^5 NC^{-1}$$
Question 4
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Coulomb's Law is true for

A
atomic distances ($$={10}^{-11}m$$)

B
nuclear distances ($$={10}^{-15}m$$)

C
charged as well as uncharged particles

D
all the distances

Solution

Coulomb's law is true for all distances whether it is small and large. Hence it is called a long range force.
Coulomb's force $$F = \dfrac{q_1 q_2}{4\pi \epsilon_o r^2}$$
$$\implies$$ $$F \propto \dfrac{1}{r^2}$$
Question 5
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In a region of space, the electric field is given by $$\vec{E}=8\hat{i}+4\hat{j}+3\hat{k}$$. The electric flux through a surface of area $$100$$ units in the xy plane is :

A
$$800$$ units

B
$$300$$ units

C
$$400$$ units

D
$$1500$$ units

Solution

$$\vec{E}=8\hat{i}+4\hat{j}+3\hat{k},\vec{A}=100\hat{k}$$
S0 $$\phi=\vec{E}\cdot \vec{A}=300 units $$
Question 6
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Two charges are placed at a certain distance apart in air. If a glass slab is introduced between them, the net force between them will :

A
increase

B
decrease

C
zero

D
cannot say

Solution

$${ F }_{ glass }=\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }{ r }^{ 2 } } =\cfrac { { F }_{ air } }{ K } $$

As $$K$$ (for glass)$$>1$$. $$\therefore {F}_{glass}<{F}_{air}$$
Question 7
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Directions For Questions

A cube has sides of length L=0.300 m. It is placed with one corner at the origin. The electric field is not uniform but is given by $$\vec{E}=(-5.00 NC^{-1})x\hat{i}+(3.00 NC^{-1})z\hat{k}$$.

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The surface that have zero flux are

A
$$S_2, S_4, $$ and $$S_5$$

B
$$S_1, S_3, S_4 $$ and $$S_6$$

C
$$S_1, S_2,$$ and $$S_3$$

D
$$S_2, S_3, $$ and $$ S_4$$

Solution

$$\vec{E} $$ lies along the x-z plane. therefore surface $$S_1$$ and $$ S_3 $$ have zero flux as both these surface lie on the x-z plane and hence no field lines component cross them perpendicularly.
At surface $$S_6$$ the magnitude of field $$\vec{E}$$ is zero, hence flux is zero.
At surface $$S_4$$ the z component of field is zero. Therefore flux is zero
Question 8
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A cube of side $$a$$ is placed such that the nearest face which is parallel to the $$y - z$$ plane is at distance $$'a'$$ from the origin. The electric field components are $$E_{x} = \alpha x^{1/2}, E_{y} = E_{z} = 0$$.
The flux $$\phi_{E}$$ through the cube is

A
$$\sqrt {2} \alpha a^{5/2}$$

B
$$-\alpha a^{5/2}$$

C
$$(\sqrt {2} - 1) \alpha a^{5/2}$$

D
Zero

Solution

Given that nearest distance of cube from origin is a,then
Flux from the nearest surface $$ \phi _1=-E a^2=-\alpha ^{1/2} a^2=-\alpha a^{5/2}$$
Flux from surface x=2a $$ \phi _2=E a^2=-\alpha(2a) ^{1/2} a^2=\sqrt 2 \alpha a^{5/2}$$
In y and z direction, there is no electric field present. So flux will be zero in that direction
$$ \phi _{net2}=(\sqrt 2-1)\alpha a^{5/2}$$
Question 9
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A certain charge $$Q$$ is divided into two parts $$q$$ and $$Q- q$$, which are then separated by a certain distance.What must $$q$$ be in terms of $$Q$$ to maximize the electrostatic repulsion between the two charges?

A
$$\displaystyle q = -\dfrac{Q}{2}$$

B
$$\displaystyle q = \dfrac{Q}{2}$$

C
$$\displaystyle q = -Q$$

D
$$0$$

Solution

Electrostatic repulsion = $$F=\dfrac{1}{4 \pi \epsilon } \dfrac{q (Q-q)}{r^2 } $$

$$ \dfrac{dF}{dq} =\dfrac{1}{4 \pi \epsilon } \dfrac{1}{r^2 } (Q - 2q) =0 \Rightarrow q =\dfrac{Q}{2} $$
Question 10
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The $$E-r$$ curve for an infinite linear charge distribution will be :

A

B

C

D

Solution

The field due to infinite linear charge distribution
$$E=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int { \cfrac { dq }{ r } } \Rightarrow \quad E\propto \cfrac { 1 }{ r } $$
So, hyperbola