Electric Charges and Fields Test

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Electric Charges and Fields Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The force between two charges when placed in free space is 5 N.If they are in a medium of relative permittivity 5, the force between them willbe

A
1 N

B
25 N

C
2.5 N

D
$$\displaystyle \frac{1}{5}$$ N

Solution

Fe in free surface
$$Fe=\quad \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { { q }^{ 2 } }{ { r }^{ 2 } } =5N$$
In medium
$$F{ e }_{ m }=\quad \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 }{ \varepsilon }_{ r } } \frac { { q }^{ 2 } }{ { r }^{ 2 } } $$
$$F{ e }_{ m }=\quad \dfrac { F{ e } }{ { \varepsilon }_{ r } } =\quad \dfrac { 5N }{ 5 } =1N$$

Option A is correct.
Question 2
1 / -0

Two like charges are placed a certain distance apart in air. If a brass plate is introduced between them, the force between the two charges will

A
increase

B
decrease

C
remain the same

D
become zero

Solution

They are like charges so they will repel each other.
If a metal plate is in between them, then its electrons will get pushed away if the two charges are negative and pulled if positive.
In both the cases, the brass plate's electrons undergo forces in opposite directions. Hence they will cancel each other and nothing will happen to the electrons.
An unaffected plate when placed between two charges doesn't affect the force between the two charges.
Question 3
1 / -0

Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will

A
increase

B
decrease

C
remain the same

D
become zero

Solution

Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will decreases.
$$F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { { q }^{ 2 } }{ { r }^{ 2 } } \quad \quad \quad \quad (in\quad vacuum)$$

But if medium is inserted:
$$F'=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 }{ \varepsilon }_{ m } } \dfrac { { q }^{ 2 } }{ { r }^{ 2 } } \quad \quad \quad \quad (in\quad medium)\\ F'=\dfrac { F }{ { \varepsilon }_{ m } } $$
Question 4
1 / -0

Two charges of 10 C and -15 C are separated in air by 1m.The ratio of force exerted by one on the other is

A
1:2

B
2:1

C
1: 1

D
none of these

Solution

From Coulomb's law, the forces exerted by one on the other are equal and opposite direction i.e $$|F_{12}|=|-F_{21}|$$
Thus the required ration will be $$1:1$$
Question 5
1 / -0

Calculate the excess number of electrons on each of two similar charged spheres that are located 5 cm apart (in air),such that the force of repulsion between them is 36 x 10$$^{-19}$$N

A
3262

B
1625

C
6250

D
1547

Solution

Using Coulomb's Law,
$$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q^{2}}{r^{2}}$$
We get, $$q^{2}=\dfrac{36\times 10^{-19}}{9\times 10^{9}}\times (5\times 10^{-2})^{2}$$
$$q^{2}=\dfrac{36}{9}\times (5)^{2}\times 10^{-32}$$
$$q=\dfrac{6}{3}\times 5\times 10^{-16}$$
$$=10\times 10^{-16}$$
$$=10^{-15}\; C$$

As $$q=ne$$
$$n=\dfrac{10^{-15}}{1.6\times 10^{-19}}$$
$$=0.625\times 10^{4}$$
$$=6250$$ electrons.
Question 6
1 / -0

An electrostatic force of attraction between two point charges $$A$$ and $$B$$ is$$ 1000 N$$. If the charge on $$A$$ is increased by $$25$$% and that on $$B$$ is reduced by $$25$$% and the initial distance between them is decreased by $$25$$%, the new force of attraction between them is ______$$N.$$

A
$$1666.67$$

B
$$3256.33$$

C
$$1253.45$$

D
$$3333.3$$

Solution

Given, $$1000=\dfrac{q_{A}q_{B}\;\times k}{r^{2}}$$ ...(i)
Now, $$F=\dfrac{\left (q_{A}\times \dfrac{5}{4} \right )\left (q_{B}\times \dfrac{3}{4} \right ) \;\times k}{r^{2}\left ( \dfrac{3}{4} \right )^{2}}$$ ...(ii)

Dividing (ii) by (i), we get,
$$\dfrac{F}{1000}=\dfrac{\dfrac{5}{4}\times\dfrac{3}{4}}{\dfrac{9}{16}}$$
$$F=1000\times \dfrac{15\times 16}{16\times 9}$$
$$F=1666.67\; N$$
Question 7
1 / -0

Two charges are placed a certain distance apart in air. If a brass plate is introduced between them, the force between them will

A
increase

B
decrease

C
remain the same

D
become zero

Solution

If two like charges are placed in air, then they will experience a force of repulsion from each other.
If a metal plate is introduced between them, then nothing will happen.
Because, if they are positively charged, then both will try to attract electrons of the plate along opposite sides and thus no net force will act on the electrons of the plate and the plate will remain as it is.
If the charges are negative then both will try to repel electrons from both sides and nothing will happen.
An unaffected plate doesn't affect the force between two charges.
Question 8
1 / -0

Two charged bodies with a distance 'd' between them are placed in water and then in the air, then the new force between them ____. (Dielectric constant of water is more than one)

A
increases

B
decreases

C
remains the same

D
first decreases then increases

Solution

$$F_{air}=\cfrac{1}{4\pi{\epsilon}_0}\cfrac{q_1q_2}{r^2}$$
$$F_{water}=\cfrac{1}{4\pi{\epsilon}}\cfrac{q_1q_2}{r^2}$$ $$[\varepsilon\longrightarrow $$Permitivity of water$$]$$
$$\varepsilon=k\varepsilon_0$$
$$\therefore{F_{air}}=KF_{water}$$
$$\therefore{F_{air}}>F_{water}$$
Question 9
1 / -0

Each of the two point charges are doubled and their distance is halved. Force of interaction becomes P times, where P is

A
1

B
4

C
$$\displaystyle \frac{1}{16}$$

D
16

Solution

By Coulomb's law, force of interaction between two charges $$q_1$$ and $$q_2$$ is $$F=\dfrac{kq_1q_2}{r^2}$$ where r be the separation between them.
When $$q_1=2q_1, q_2=2q_2, r=r/2$$, the force interaction becomes $$F'=\dfrac{k(2q_1)(2q_2)}{(r/2)^2}=16F$$
Thus, the value of P will be $$16$$.
Question 10
1 / -0

Two identical spheres having charges $$8 \mu C$$ and $$-4\mu C$$ are kept at a certain distance apart. Now they are brought into contact, after that again they are kept at the same distance. Compare the forces in the two cases

A
$$4 : 1$$

B
$$12 : 1$$

C
$$8 : 1$$

D
$$16 : 1$$

Solution

If two identical sphere having charge $$8\: \mu C$$ and $$-4\: \mu C$$ are kept a certain distance apart, then force between them is $$F$$.
Let the distance between charged sphere is $$d$$.
then, $$F=\dfrac{k\times 8\times (-4)\times 10^{-12}}{d^{2}}$$ [attractive]
$$d^{2}=\dfrac{32k\times 10^{12}}{F}$$ ....(i)

Now, when both the identical sphere touch, then charge on each sphere is average of initial charge on sphere.
i.e, $$Q=\dfrac{(8-4)\:\mu C}{2}= 2\; \mu C$$ on each sphere.
Now, Force between them is:
$$F'=\dfrac{k\times (2\times 10^{-6})^{2}}{d^{2}}$$
From equation (i), we get
$$F'=\dfrac{k\times 4\times 10^{-12}\times F}{32k\times 10^{-12}}$$
$$F'=\dfrac{F}{8}$$
so, the ratio is $$8:1$$

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Electric Charges and Fields Test - 38

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Electric Charges and Fields Test - 38

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Question 1

1 N

25 N

2.5 N

$$\displaystyle \frac{1}{5}$$ N

Solution

Question 2

increase

decrease

remain the same

become zero

Solution

Question 3

increase

decrease

remain the same

become zero

Solution

Question 4

1:2

2:1

1: 1

none of these

Solution

Question 5

3262

1625

6250

1547

Solution

Question 6

$$1666.67$$

$$3256.33$$

$$1253.45$$

$$3333.3$$

Solution

Question 7

increase

decrease

remain the same

become zero

Solution

Question 8

increases

decreases

remains the same

first decreases then increases

Solution

Question 9

1

4

$$\displaystyle \frac{1}{16}$$

16

Solution

Question 10

$$4 : 1$$

$$12 : 1$$

$$8 : 1$$

$$16 : 1$$

Solution

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