Electric Charges and Fields Test

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Electric Charges and Fields Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The number of electrons present in $$1\ C$$ of charge is ______.

A
$$4.25\times 10^{18}$$

B
$$6.25\times 10^{18}$$

C
$$3.25\times 10^{18}$$

D
$$2.25\times 10^{18}$$

Solution

Charge possessed by one electron is $$e=1.6\times 10^{-19}C$$

$$Q=1C$$

Number of electrons =$$N$$

$$Q=Ne$$

$$\implies N=\dfrac{Q}{e}=\dfrac{1}{1.6\times 10^{-19}}$$

$$\implies N=6.25\times 10^{18}$$

Answer-(B)
Question 2
1 / -0

Two charged particles having charges $$+25\mu C$$ and $$+50\mu C$$ are separated by a distance of $$8\ cm$$. The force acting on them is:

A
$$50 kN$$

B
$$10kN$$

C
$$2.7 kN$$

D
$$1.7 kN$$

Solution

$$Force = \cfrac{1}{4\pi {\epsilon}_0} \cfrac{q_1 q_2}{r^2}$$
$$= \cfrac{9 \times {10}^9 \times 25 \times {10}^{-6} \times 50 \times {10}^{-6}}{(0.08)^2}$$
$$= \cfrac{9 \times 25 \times 5}{64} \times {10}^2 \; N$$
$$= 1.7 \times {10}^3 \; N$$
$$= 1.7 \; KN$$
Question 3
1 / -0

If two charges of $$1$$ coulomb each are placed $$1\ km$$ apart in vacuum, the force between them will be:

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

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

C
$$1.1\times 10^{-4}N$$

D
$$10^{-6}N$$

Solution

$$Q=1C$$ $$r=1000m$$
$$F=\dfrac{KQQ}{r^2}=\dfrac{9\times 10^9}{((10)^3)^2}=9\times10^3N$$
Question 4
1 / -0

A charge of $$-30\mu C$$ is placed at $$50\ cm$$ from another charge of $$+25\mu C$$ in vacuum. The force between them is:

A
$$60\ N$$

B
$$27\ N$$

C
$$90\ N$$

D
$$120\ N$$

Solution

$$q_1=-30\times 10^{-6}$$, $$q_2=25\times 10^{-6}$$ and $$r=.5$$
$$F=\dfrac{kq_1q_2}{r^2}=27 N$$
Question 5
1 / -0

A point charge q produces an electric field of magnitude $$2\ N\ C^{-1}$$ at a point distance $$0.25\ m$$ from it. Find the value of charge.

A
$$13.9\times 10^{-10}C$$

B
$$13.9\times 10^{-21}C$$

C
$$13.9\times 10^{-11}C$$

D
$$13.9\times 10^{-12}C$$

Solution

electric field,
$$ E=\cfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \cfrac { q }{ { r }^{ 2 } } $$
given,$$E=2N{ c }^{ -1 }\\ r=0.25m\\ so,q=E\times 4\pi { \varepsilon }_{ o }\times { r }^{ 2 }\\ =2\times \cfrac { 1 }{ 9\times { 10 }^{ 9 } } \times { (0.25) }^{ 2 }\\ q=0.01388\times { 10 }^{ -9 }\\ =13.9\times { 10 }^{ -12 }C$$
so; the correct answer option is d.
Question 6
1 / -0

Two charged particles having charge $$2\times 10^{-8}C$$ each are joined by an insulating string of length $$100\ cm$$. The system is kept on a horizontal table. Find the tension in the string.

A
$$3.6\times 10^{-7}N$$

B
$$3.6\times 10^{-6}N$$

C
$$3.6\times 10^{-4}N$$

D
$$3.6\times 10^{-5}N$$

Solution

Tension in string is equal to the force by which two charges repel each other.

$$q_1=q_2=2\times 10^{-8}C$$ and $$r=100cm=1m$$

$$T=F_e=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}$$

$$\implies T=9\times 10^9\times \dfrac{4\times 10^{-16}}{1^2}$$

$$\implies T=36\times 10^{-7}N$$

$$\implies T=3.6\times 10^{-6}N$$

Answer-(B)
Question 7
1 / -0

Two particles having charges $$q_{1}$$ and $$q_{2}$$ when kept at a certain distance, exert a force F on each other. If the distance between the two charges is reduced to half then the force exerted on each other would be:

A
$$2\ F$$

B
$$4\ F$$

C
$$8\ F$$

D
$$16\ F$$

Solution

Let the force be $$F=\dfrac{kq_1q_2}{r^2}$$ ....(1)
so distance is reduced to $$r/2$$
so $$F_1=\dfrac{kq_1q_2}{(\frac{r}{2})^2}=\dfrac{4kq_1q_2}{r^2}=4F$$ by using equation 1.
Question 8
1 / -0

When a glass rod is rubbed with a piece of silk cloth the rod ?

A
And the cloth both acquire positive charge.

B
Becomes positively charged while the cloth has a negative charge.

C
And the cloth both acquire negative charge.

D
Becomes negatively charged while the cloth has a positive charge.

Solution

It is a convention to call the charge acquired by a glass rod when it is rubbed with silk as positive. The other kind of charge is said to be negative
Question 9
1 / -0

The electrostatic force between two point charges $$q_{1}$$ and $$q_{2}$$ at separation 'r' is given by $$F = \dfrac {Kq_{1}q_{2}}{r^{2}}$$. The constant K:

A
Depends on the system of units only

B
Depends on the medium between the charges only

C
Depends on both the system of units and the medium between the charges

D
Is independent of both the system of units and the medium between the charges

Solution

$$K=\dfrac{1}{4\pi\epsilon_ok}$$ and its units is $$N-m^2/C^2$$
It depend on system of unit and dielectric constant (k), that's medium between charges
Question 10
1 / -0

A charged particle of mass $${m}_{1}$$ and charge $${q}_{1}$$ is revolving in a circle of radius $$r$$. Another charged particle of charge $${q}_{2}$$ and mass $${m}_{2}$$ is situated at the centre of the circle. If the velocity and time period of the revolving particle be $$v$$ and $$T$$ respectively, then :

A
$$v=\sqrt { \cfrac { { q }_{ 1 }{ q }_{ 2 }r }{ 4\pi { \epsilon }_{ 0 }{ m }_{ 1 } } } $$

B
$$v=\cfrac { 1 }{ { m }_{ 1 } } \sqrt { \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }r } } $$

C
$$T=\sqrt { \cfrac { 16{ \pi }^{ 3 }{ \epsilon }_{ 0 }{ m }_{ 1 }^{ 2 }{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

D
$$T=\sqrt { \cfrac { 16{ \pi }^{ 3 }{ \epsilon }_{ 0 }{ m }_{ 2 }^{ }{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

E
None of the above

Solution

Both charges are either both positive or both negative(since answers has the product $$q_1q_2$$ inside square root). Hence, circular motion is not possible. So either $$q_1$$ or $$q_2$$ should be negative.

According to the diagram,
$$\dfrac{m_1v^2}{r}=\dfrac{1}{4\pi \epsilon_0}\dfrac{q_1q_2}{r^2}$$
$$\implies v=\sqrt{\dfrac{1}{4\pi\epsilon_0m_1}\dfrac{q_1q_2}{r}}$$
On putting, $$v=r\omega$$ and $$\omega=\dfrac{2\pi}{T}$$, we can find out:
$$T=\sqrt{\dfrac{16\pi^3\epsilon_0m_1r^3}{q_1q_2}}$$
Hence, none of the answers is correct.

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Electric Charges and Fields Test - 43

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Electric Charges and Fields Test - 43

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Question 1

$$4.25\times 10^{18}$$

$$6.25\times 10^{18}$$

$$3.25\times 10^{18}$$

$$2.25\times 10^{18}$$

Solution

Question 2

$$50 kN$$

$$10kN$$

$$2.7 kN$$

$$1.7 kN$$

Solution

Question 3

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

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

$$1.1\times 10^{-4}N$$

$$10^{-6}N$$

Solution

Question 4

$$60\ N$$

$$27\ N$$

$$90\ N$$

$$120\ N$$

Solution

Question 5

$$13.9\times 10^{-10}C$$

$$13.9\times 10^{-21}C$$

$$13.9\times 10^{-11}C$$

$$13.9\times 10^{-12}C$$

Solution

Question 6

$$3.6\times 10^{-7}N$$

$$3.6\times 10^{-6}N$$

$$3.6\times 10^{-4}N$$

$$3.6\times 10^{-5}N$$

Solution

Question 7

$$2\ F$$

$$4\ F$$

$$8\ F$$

$$16\ F$$

Solution

Question 8

And the cloth both acquire positive charge.

Becomes positively charged while the cloth has a negative charge.

And the cloth both acquire negative charge.

Becomes negatively charged while the cloth has a positive charge.

Solution

Question 9

Depends on the system of units only

Depends on the medium between the charges only

Depends on both the system of units and the medium between the charges

Is independent of both the system of units and the medium between the charges

Solution

Question 10

$$v=\sqrt { \cfrac { { q }_{ 1 }{ q }_{ 2 }r }{ 4\pi { \epsilon }_{ 0 }{ m }_{ 1 } } } $$

$$v=\cfrac { 1 }{ { m }_{ 1 } } \sqrt { \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \epsilon }_{ 0 }r } } $$

$$T=\sqrt { \cfrac { 16{ \pi }^{ 3 }{ \epsilon }_{ 0 }{ m }_{ 1 }^{ 2 }{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

$$T=\sqrt { \cfrac { 16{ \pi }^{ 3 }{ \epsilon }_{ 0 }{ m }_{ 2 }^{ }{ r }^{ 3 } }{ { q }_{ 1 }{ q }_{ 2 } } } $$

None of the above

Solution

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