Electric Charges and Fields Test

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Electric Charges and Fields Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The force of interaction (attraction or repulsion) between two stationary point charges in vacuum is directly proportional to the product of the charges and inversely proportional to the square of distance between them. This is called_______

A
Curie's Law

B
Coulomb's law

C
Faraday's Law

D
Ampere's Law

Solution

The given statement is of Coulomb;s law which states that force between two stationary point charges is $$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{r^{2}}$$
where $$q_{1}$$ and $$q_{2}$$ are two stationary point charges and $$r$$ is the distance between them.
$$\dfrac{1}{4\pi \varepsilon _{0}}=$$ constant of prportionality.
Question 2
1 / -0

If the electric field is given by $$(5i+4j+9k)$$, the electric flux through a surface of area $$20$$ unit lying in the Y-Z plane will be :

A
$$100$$ unit

B
$$80$$ unit

C
$$180$$ unit

D
$$20$$ unit

Solution

As surface of area lies in the Y-Z plane, thus its area vector points in X direction i.e. $$\vec{A} = 20\hat{i}$$
Electric flux $$\phi = \vec{E}.\vec{A}$$
$$\therefore$$ $$\phi = (5\hat{i}+4\hat{j}+9\hat{k}).(20\hat{i})$$
$$\implies $$ $$\phi = 5\times 20 = 100$$ unit
Question 3
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The force of interaction (attraction or repulsion) between two stationary point charges in vacuum is _____

A
directly proportional to the product of the charges

B
inversely proportional to the product of the charges

C
directly proportional to only the lower of the charges

D
inversely proportional to only the lower of the charges

Solution

Electrostatic force=$$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}$$

Hence, force of interaction is directly proportional to the product of charges.

Answer-(A)
Question 4
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Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of dielectric constant $$4$$ is introduced between them, the force will be:

A
$$4\ F$$

B
$$2\ F$$

C
$$F/2$$

D
$$F/4$$

Solution

$$\textbf{Step 1: Force between two charges in vaccum}$$
Applying Coulomb's law,
$$F = \dfrac{1}{4\pi \in_0} \dfrac{q_1q_2}{d^2} $$ $$ ....(1)$$
where $$d$$ is the distance between the charges

$$\textbf{Step 2: Force between charges in medium}$$
Ler $$k$$ be the dielectric constant of the medium.
$$F' = \dfrac{1}{4\pi \in_0} \dfrac{ 1}{k} \dfrac{q_1q_2}{d^2}$$

$$\Rightarrow F ' = \dfrac{F}{k}=\dfrac{F}{4}$$

Hence, option (D) is correct.
Question 5
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Two spheres carrying charges $$+ 6\mu C$$ and $$+ 9\mu C$$ separated by a distance $$d$$, experiences a force of repulsion $$F$$. When a charge of $$- 3\mu C$$ is given to both the sphere and kept at the same distance as before, the new force of repulsion is :

A
$$F$$

B
$$3F$$

C
$$F/3$$

D
$$F/9$$

Solution

$$\textbf{Step 1: Coulomb's law in initial situation [Refer Fig. 1]} $$
$$Q_{1} = +6\ \mu C ,\ \ Q_{2} = +9\ \mu C $$

$$F = \dfrac{K Q_1 Q_2}{r^2} $$ $$ = \dfrac{K(6\ \mu C)(9\ \mu C)}{d^2} $$ $$....(1)$$

$$\textbf{Step 2: Final charge distribution when}\, - 3\ \mu C\,\, \textbf{charge added to both the sphere.}$$ $$\textbf{[Refer Fig. 2]} $$

$$\textbf{Step 3: Apply Coulomb's law in final Situation [Refer Fig. 3]} $$
Distance between the spheres remains same $$=d$$.

$$F_{1} = \dfrac{KQ_{1}' Q_{2}'}{d^2} $$ $$= \dfrac{K(3\ \mu C)(6\ \mu C)}{d^2} $$ $$....(2)$$

Divide equation $$(2)$$ by equation $$(1)$$

$$\dfrac{F_1}{F} = \dfrac{6 \times 3}{6 \times 9}\ \ \Rightarrow\ \ F_{1} = \dfrac{F}{3} $$

Hence force in final situation is equal to $$ \dfrac{F}{3} $$
Option $$(C)$$ correct
Question 6
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Two point charges having equal charges separated by $$1 m$$ distance experience a force of $$8 N$$. What will be the force experienced by them, if they are held in water, at the same distance?
(Given, $$K_{water} = 80$$ )

A
$$1N$$

B
$$0.01N$$

C
$$0.1N$$

D
$$10N$$

Solution

$$F=\dfrac{q_1q_2}{4\pi\epsilon_o K r^2}$$

For air, $$K=1$$ and hence $$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}=8N$$

In water:-

$$F=\dfrac{q_1q_2}{4\pi\epsilon_o K r^2}=\dfrac{8N}{k}=\dfrac{8}{80}$$

$$\implies F=0.1N$$

Answer-(C)
Question 7
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Coulomb's Law agrees with________

A
Newtons $$3$$rd Law of Motion

B
Newtons $$1$$st Law of Motion

C
Newtons $$2$$nd Law of Motion

D
All of the above

Solution

Coulomb's law agrees with Newton's $$1st$$, $$2nd$$ and $$3rd$$ law of motion. Coulomb's law states that $$F=k\dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } $$ here the two charge particle is similar to the objects of the Newton's $$1st$$ law. Newton's $$3rd$$ law is same as the two similar charge particles repel to each other. Which coincides with Newton's $$3rd$$ law every action there is an equal and opposite reaction.
Question 8
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The sum of two point charges is $$7 \mu C$$. They repel each other with a force of $$1 N$$ when kept $$30 cm$$ apart in free space. Calculate the value of each charge.

A
$$1\mu C$$ and $$6 \mu C$$

B
$$2\mu C$$ and $$-5 \mu C$$

C
$$3\mu C$$ and $$4 \mu C$$

D
$$2\mu C$$ and $$5 \mu C$$

Solution

$$q_1+q_2=7\mu C$$

$$F=\dfrac{kq_1q_2}{r^2}$$

$$\implies \dfrac{9\times 10^9\times q_1q_2}{(0.3)^2}=1N$$

$$\implies q_1q_2=10^{-11}=10 (\mu C)^2$$

Now, $$q_1-q_2=\sqrt{(q_1+q_2)^2-4q_1q_2}$$

$$\implies q_1-q_2=3\mu C$$

And, $$q_1+q_2=7\mu C$$

Hence, $$q_1=5\mu C$$ and $$q_2=2\mu C$$

Answer-(D)
Question 9
1 / -0

Coulomb's law is a confirmation of______

A
inverse cube law

B
product law

C
inverse square law

D
None of the above

Solution

According to Coulomb's Law-

$$F\propto \dfrac{1}{r^2}$$

Hence, Coulomb's Law is a confirmation of inverse square law.

Answer-(C)
Question 10
1 / -0

Charges of $$1 \mu C$$ and $$10\mu C$$ are placed and held $$0.1m$$ apart. What will be the force they exert on each other?

A
$$0.9N$$

B
$$90N$$

C
$$9N$$

D
$$900N$$

Solution

$$F=\dfrac{kq_1q_2}{r^2}$$

$$\implies F=\dfrac{9\times 10^9\times 10^{-6}\times 10\times 10^{-6}}{(0.1)^2}$$

$$\implies F=9N$$

Answer-(C)

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Electric Charges and Fields Test - 44

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Electric Charges and Fields Test - 44

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Question 1

Curie's Law

Coulomb's law

Faraday's Law

Ampere's Law

Solution

Question 2

$$100$$ unit

$$80$$ unit

$$180$$ unit

$$20$$ unit

Solution

Question 3

directly proportional to the product of the charges

inversely proportional to the product of the charges

directly proportional to only the lower of the charges

inversely proportional to only the lower of the charges

Solution

Question 4

$$4\ F$$

$$2\ F$$

$$F/2$$

$$F/4$$

Solution

Question 5

$$F$$

$$3F$$

$$F/3$$

$$F/9$$

Solution

Question 6

$$1N$$

$$0.01N$$

$$0.1N$$

$$10N$$

Solution

Question 7

Newtons $$3$$rd Law of Motion

Newtons $$1$$st Law of Motion

Newtons $$2$$nd Law of Motion

All of the above

Solution

Question 8

$$1\mu C$$ and $$6 \mu C$$

$$2\mu C$$ and $$-5 \mu C$$

$$3\mu C$$ and $$4 \mu C$$

$$2\mu C$$ and $$5 \mu C$$

Solution

Question 9

inverse cube law

product law

inverse square law

None of the above

Solution

Question 10

$$0.9N$$

$$90N$$

$$9N$$

$$900N$$

Solution

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