Electric Charges and Fields Test

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Electric Charges and Fields Test - 40

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Weekly Quiz Competition

Question 1
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A large sphere of charge +Q is fixed in position. A smaller sphere of charge +q is placed near the larger sphere and released from rest then choose the correct option, smaller sphere will move with:

A
decreasing velocity and decreasing acceleration

B
decreasing velocity and increasing acceleration

C
decreasing velocity and constant acceleration

D
increasing velocity and decreasing acceleration

E
increasing velocity and increasing acceleration

Solution

Force between charges at distance $$r=\dfrac{kQq}{r^2}$$. This force is repulsive in nature. Thus the spheres tend to move away. Since the force remains positive for all values of $$r$$, the velocity is always increasing.
However the magnitude of force decreases with increasing distance between charges $$r$$. Thus the acceleration keeps on decreasing.
Question 2
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Two particles A & B having charges $$+q$$ and $$-2q$$. What happens to Electrostatic force between them if the distance between the A & B is halved?

A
It decreases by a factor of $$4$$

B
It decreases by a factor of $$2$$

C
It remains the same

D
It increases by a factor of $$2$$

E
It increases by a factor of $$4$$

Solution

The electrostatic force between two charges is $$F=\dfrac{k(q)(-2q)}{r^2}$$
When distance r is halved, the force $$F'=\dfrac{k(q)(-2q)}{(r/2)^2}=4F$$
Question 3
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Two charged spheres are kept apart by $$2 mm$$. Identify which of the following set of charges would yield the greatest attractive force between two spheres?

A
$$+1q$$ and $$+4q$$

B
$$-1q$$ and $$-4q$$

C
$$+2q$$ and $$+2q$$

D
$$-2q$$ and $$-2q$$

E
$$+2q$$ and $$-2q$$

Solution

The spheres must have unlike charges so that the force between them is an attractive force.
Coulomb's force $$F = \dfrac{k Q_1 Q_2}{r^2}$$ where $$k = \dfrac{1}{4\pi \epsilon_o} =constant$$
$$\implies$$ $$F \propto Q_1 Q_2$$
Thus the spheres with charge $$+2q$$ and $$-2q$$ will experience the greatest attractive force.
Question 4
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What happens to the force between two charged particles when they are moved $$50$$% closer to one another?

A
The force between them doubles

B
The force between them halves

C
The force between them remains the same

D
The force between them reverses

E
The force between them operates in the same direction

Solution

Let the initial distance between the two charges be $$r$$.
$$\therefore$$ Force between them $$F =\dfrac{KQ_1Q_1}{r^2}$$
Now they are moved closer, thus new distance between them $$r' = \dfrac{r}{2}$$
New force $$F' = \dfrac{KQ_1Q_2}{(r/2)^2} = 4F$$
Thus the force between them becomes $$4$$ times the initial force.

Force between two charges always act along the line of the two charges. Hence, the direction of force does not change.
Question 5
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Which of the following is true when the distance between two point charges is halved?

A
The particles attract one another

B
The particles repel one another

C
The particles exert the same force on one another

D
The particles exert twice as much force on one another

E
The particles exert four times as much force on one another

Solution

The force between two point charges $${q}_{1}$$ and $${q}_{2}$$ separated by a distance $$r$$ is given by:
$$E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{{q}_{1}{q}_{2}}{{r}^{2}}$$

When the distance is halved, then electric force will be
$$E'=\dfrac{1}{4\pi\varepsilon_0}\dfrac{{q}_{1}{q}_{2}}{{\left({{\frac{r}{2}}}\right)}^{2}}$$

Now dividing $$E'$$ by $$E$$ we get
$$\dfrac{E'}{E}=4$$
or $$E'=4E$$
therefore force is four times greater now.
Question 6
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Two point charges $$+8\mu C$$ and $$+12\mu C$$ repel each other with a force of $$48\ N$$. When an additional charge of $$-10\mu C$$ is given to each of these charges (the distance between the charges is unaltered), then the new force is:

A
repulsive force of $$24\ N$$

B
attractive force of $$24\ N$$

C
repulsive force of $$2\ N$$

D
attractive force of $$2\ N$$

Solution

$$q_1=+8\ \mu C$$

$$q_2=+12\ \mu C$$

$$F_1=48\ N$$

Now, $$q'_1=-2\ \mu C$$

and, $$q'_2=+2\ \mu C$$

So, $$\dfrac{F_1}{F_2}=\dfrac{q_1q_2}{q'_1q'_2}$$

$$F_2=2N$$
Question 7
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A Gaussian surface in the cylinder of cross section $$\displaystyle \pi { a }^{ 2 }$$ and length $$L$$ is immersed in a uniform electric field $$\displaystyle \overline { E } $$ with the cylinder axis parallel to the field. The flux $$\displaystyle '\phi '$$ of the electric field through the closed surface is:

A
$$\displaystyle 2\pi { a }^{ 2 }\overline { E } $$

B
$$\displaystyle \pi { a }^{ 2 }\overline { E } L$$

C
$$\displaystyle \pi { a }^{ 2 }\left( 2+L \right) \overline { E } $$

D
Zero

Solution

The electric lines of force would enter from one surface of the cylinder and leave out from the other.
Thus the net flux through the cylinder would be :
$$\vec{B}.\vec{A_1}+\vec{B}.\vec{A_2}$$
$$B(\pi a^2)+B(-\pi a^2)=0$$
Question 8
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The electric force between two point charges separated by a certain distance in air is $$F$$ the distance at which they should be placed in a medium of relative permittivity $$k$$ so that the force remain the same is:

A
$$d$$

B
$$\dfrac { d }{ k } $$

C
$$kd$$

D
$$\dfrac { d }{ \sqrt { k } } $$

Solution

$${ F }_{ m }={ F }_{ a }$$

$$\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 }k } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { d_m }^{ 2 } } =\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { d }_{ a }^{ 2 } } $$

$$\Rightarrow { d }_{ a }^{ 2 }=k{ d }_{ m }^{ 2 }$$

$$\Rightarrow { d }_{ m }=\dfrac { d }{ \sqrt { k } } $$
Question 9
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The force between two point charges is F, which of the following could be done to make the force between them 1/2 F?

A
Reduce the charge on each point charge to half of its original value and double the distance between the point charges.

B
Increase the charge on each point charge to double its original value and double the distance between them.

C
Increase the charge on each point charge to double its original value and move the point charges so that only half the distance is between them.

D
Increase the charge on only one point charge to twice its value and double the distance between the point charges.

Solution

By Coulomb's law, $$F=\dfrac{kq_1q_2}{r^2}$$
For A : $$F'=\dfrac{k(q_1/2)(q_2/2)}{(2r)^2}=F/16$$
For B : $$F'=\dfrac{k(2q_1)(2q_2)}{(2r)^2}=F$$
For C : $$F'=\dfrac{k(2q_1)(2q_2)}{(r/2)^2}=16F$$
For D : $$F'=\dfrac{k(2q_1)(q_2)}{(2r)^2}=F/2$$
Thus, option D will correct.
Question 10
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Two masses $${M}_{1}$$ and $${M}_{2}$$, which have a charge $${Q}_{1}$$ and $${Q}_{2}$$, respectively are placed at a distance $$r$$ initially.
If there is no other force acting on the charged particles and they remain at a distance $$r$$ apart indefinitely, identify which of the following must be true?

A
Both charges are positive

B
$${Q}_{1}$$ is positive and $${Q}_{2}$$ is negative

C
$${Q}_{1}$$ is negative and $${Q}_{2}$$ is positive

D
$${Q}_{1} = {Q}_{2}$$

E
$${M}_{1} = {M}_{2}$$

Solution

The gravitational force between the two masses is given as $$-\dfrac{GM_1M_2}{r^2}$$
and the electric force between them is given as $$\dfrac{1}{4\pi\epsilon_0}\dfrac{Q_1Q_2}{r^2}$$.
For the masses to remain stationary under these two forces, the net force on each must be zero.
Thus $$-\dfrac{GM_1M_2}{r^2}+\dfrac{1}{4\pi\epsilon_0}\dfrac{Q_1Q_2}{r^2}=0$$
$$\implies Q_1Q_2>0$$
Thus they must have same sign.
Hence correct answer is option A.

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Electric Charges and Fields Test - 40

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Question 1

decreasing velocity and decreasing acceleration

decreasing velocity and increasing acceleration

decreasing velocity and constant acceleration

increasing velocity and decreasing acceleration

increasing velocity and increasing acceleration

Solution

Question 2

It decreases by a factor of $$4$$

It decreases by a factor of $$2$$

It remains the same

It increases by a factor of $$2$$

It increases by a factor of $$4$$

Solution

Question 3

$$+1q$$ and $$+4q$$

$$-1q$$ and $$-4q$$

$$+2q$$ and $$+2q$$

$$-2q$$ and $$-2q$$

$$+2q$$ and $$-2q$$

Solution

Question 4

The force between them doubles

The force between them halves

The force between them remains the same

The force between them reverses

The force between them operates in the same direction

Solution

Question 5

The particles attract one another

The particles repel one another

The particles exert the same force on one another

The particles exert twice as much force on one another

The particles exert four times as much force on one another

Solution

Question 6

repulsive force of $$24\ N$$

attractive force of $$24\ N$$

repulsive force of $$2\ N$$

attractive force of $$2\ N$$

Solution

Question 7

$$\displaystyle 2\pi { a }^{ 2 }\overline { E } $$

$$\displaystyle \pi { a }^{ 2 }\overline { E } L$$

$$\displaystyle \pi { a }^{ 2 }\left( 2+L \right) \overline { E } $$

Zero

Solution

Question 8

$$d$$

$$\dfrac { d }{ k } $$

$$kd$$

$$\dfrac { d }{ \sqrt { k } } $$

Solution

Question 9

Reduce the charge on each point charge to half of its original value and double the distance between the point charges.

Increase the charge on each point charge to double its original value and double the distance between them.

Increase the charge on each point charge to double its original value and move the point charges so that only half the distance is between them.

Increase the charge on only one point charge to twice its value and double the distance between the point charges.

Solution

Question 10

Both charges are positive

$${Q}_{1}$$ is positive and $${Q}_{2}$$ is negative

$${Q}_{1}$$ is negative and $${Q}_{2}$$ is positive

$${Q}_{1} = {Q}_{2}$$

$${M}_{1} = {M}_{2}$$

Solution

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