Electric Charges and Fields Test

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Electric Charges and Fields Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

A charge of $$10$$ $$units$$ is divided so that force between the two charges is maximum when placed $$2$$ $$cm$$ apart. The two charges are:

A
$$6, 4$$

B
$$3, 7$$

C
$$1, 9$$

D
$$5, 5$$

Solution

Let the two charges be $$q_1, q_2$$
$$q_1+q_2=10$$
Force between them when placed 2 cm apart
$$F=\dfrac {k(q_1)(q_2)}{(2\times 10^{-2})^2}$$
$$F=\dfrac {k(q_1)(10-q_1)}{(2\times 10^{-2})^2}$$
differentiate $$F$$ and equate it to zero to find at value of $$q_1$$, $$F$$ will be maximum.
$$\dfrac {dF}{dq_1}=\dfrac {k}{(2\times 10^{-2})^2}(10-2q_1)=0$$
$$\Rightarrow q_1=5$$
$$q_2=10-q_1$$
$$q_2=5$$
$$\therefore q_1=q_2=5\ C$$
Question 2
1 / -0

The electric field in a region of space is given by $$\vec{E}=(\hat{5i}+\hat{2j} )Nc^{-1}$$ . The electric flux due to this field through an area $$2m^{2}$$ lying in the Y-Z plane in S.I. units is

A
$$10$$

B
$$20$$

C
$$10\sqrt{2}$$

D
$$2\sqrt{29}$$

Solution

As $$2m^{2}$$ area is in y-z plane its area vector which is perpendicular to y-z plane is in x axis.
$$\therefore$$ area vector $$=2(\hat{i})$$
given $$E=5i+2j$$
We know flux $$\phi=E.A$$
$$=(2i).(5i+2j)$$
$$\phi=10$$
Question 3
1 / -0

A charge $$Q$$ is divided into two parts $$q_{1}$$ and $$q_{2}$$ such that they experience maximum force of repulsion when separated by certain distance. The ratio of $$Q, q_{1}$$ and $$q_{2}$$ is :

A
1 : 1 : 2

B
1 : 2 : 2

C
2 : 2 : 1

D
2 : 1 : 1

Solution

$$\textbf{Step 1: Divide Q} $$
$$Q$$ is divided into two parts $$q_1$$ and$$q_2$$
$$\therefore $$ $$Q = q_{1} + q_{2} $$ $$....(1)$$

$$\textbf{Step 2: Force of repulsion between}\ q_{1}\ \textbf{and}\ q_{2}$$
Let $$r$$ be the distance between charges.
Using Coulomb's law, $$F = \dfrac{k q_{1} q_{2}}{r^2} $$$$ = \dfrac{k q_{1} (Q - q_{1})}{r^2} $$ $$....(2)$$

$$\textbf{Step 3: Condition for force to be maximum} $$
The repulsive force between the charges will be maximum if,
$$\dfrac{dF}{dq_{1}} = 0 $$
$$\Rightarrow\dfrac{k}{r^2} (Q - 2q_{1}) = 0$$
$$\Rightarrow q_{1} = \dfrac{Q}{2} $$

Equation $$(1)\Rightarrow$$ $$q_{2} = Q - \dfrac{Q}{2} = \dfrac{Q}{2} $$

$$\textbf{Step 4: Ratio of}\ Q : q_{1} : q_{2} $$

$$Q : q_{1} : q_{2} = Q : \dfrac{Q}{2} : \dfrac{Q}{2} = 1 : \dfrac{1}{2} : \dfrac{1}{2} $$
$$\Rightarrow Q : q_{1} : q_{2} = 2 : 1 : 1 $$

Hence, Option (D) is correct.
Question 4
1 / -0

Two alpha particles are separated by a distance of $$10^{-13}m$$. The force between them in free space is :

A
$$92.16\times10^{-3}N $$

B
$$92.16\times10^{-4}N $$

C
$$92.16\times10^{-5}N $$

D
$$92.16\times10^{-2}N $$

Solution

$$\alpha \ particles \ are\ He^{+2} \ ions$$
$$\therefore $$ Net charge present on them is $$2\times (1\cdot 6\times 10^ {-19})C.$$
We know $$F=\dfrac {kq_1q_2}{r^2}$$

$$F=\dfrac {(9\times 10^9)(2\times 1\cdot 6\times 10^{-19})(2\times 1\cdot 6\times 10^{-19})}{(10^{-13})^2}$$

$$F=\dfrac {92\cdot 16\times 10^{-29}}{10^{-26}}$$

$$F=92\cdot 16\times 10^{-3}N$$
Question 5
1 / -0

The excess (equal in number) number of electrons that must be placed on each of the two small spheres spaced $$3 cm$$ apart with a force of repulsion between them to be $$10^{-19} N$$ is :

A
$$25$$

B
$$225$$

C
$$625$$

D
$$1250$$

Solution

We need force of repulsion $$= 10^{-19}N$$
Let n no. of increase electrons are placed on each sphere
$$\Rightarrow \dfrac{K(ne)(ne)}{(3\times 10^{-2})^2}= 10^{-19}$$

$$\Rightarrow n^2e^2= \dfrac{10^{-19}\times 9\times 10^{-4}}{9\times 10^9\times e^2}$$

$$\Rightarrow n^2= \dfrac{10^{-32}}{e^2} \Rightarrow n= \dfrac{10^{-16}}{e}$$

$$n= \dfrac{10^{-16}}{1.6\times 10^{-19}}$$

$$n= \dfrac{1000}{1.6}= 625$$
Question 6
1 / -0

Two charges each of $$100$$ micro coulomb are separated in a medium of relative permittivity $$2$$ by a distance of $$5 cm$$. The force between them is :

A
$$0.36\times10^{5}N $$

B
$$3.6\times10^{5}N $$

C
$$1.8\times10^{4}dyne $$

D
$$1.8\times10^{4}N $$

Solution

We know force between two charges in a medium separated by a distance 'r'
$$F=\dfrac {(q_1)(q_2)}{4\pi \varepsilon r^2}$$

where, $$ \epsilon_r= \dfrac{\epsilon}{\epsilon_o}$$
given, $$\epsilon_r=2$$ $$\implies \epsilon = 2\times \epsilon_o$$

$$F=\dfrac {(100\times 10^{-6})(100\times 10^{-6})\times 9\times 10^9}{2\times (5\times 10^{-2})^2}$$

$$F=\dfrac {90}{2\times 25\times 10^{-4}}$$

$$F=\dfrac {90\times 100\times 10^2}{50}$$

$$F=1\cdot 8\times 10^4 N$$
Question 7
1 / -0

Two equally charged pith balls $$3\ cm$$ apart repel each other with a force of $$4\times10^{-5}\ N $$. The charge on each ball is

A
$$2\times10^{9}\ C $$

B
$$2\times10^{-9}\ C $$

C
$$\dfrac{2}{3}\times10^{9}\ C $$

D
$$\dfrac{2}{3}\times10^{-9}\ C $$

Solution

The pith balls repel each other with force $$F=4\times 10^{-5}\ N$$ and let the charge on each pith ball be $$Q$$.

$$\therefore \dfrac {k(Q)(Q)}{(3\times 10^{-2})^2}=4\times 10^{-5}$$

or, $$Q^2=\dfrac {4\times 10^{-5}\times 9\times 10^{-4}}{9\times 10^9}$$

or, $$Q^2=4\times 10^{-18}$$

or, $$Q=2\times 10^{-9}\ C$$
Question 8
1 / -0

$$Q_{1},Q_{2}$$ charges are separated by distance $$‘d’$$. $$F$$ is the force between them. Now the value of $$Q_{2}$$ is halved. To have the same force between the charges, the distance of separation should be :

A
$$d$$

B
$$\dfrac{d}{2}$$

C
$$\dfrac{d}{\sqrt{2}}$$

D
$$2d$$

Solution

Initially force between them $$F= \dfrac{K Q_1 Q_2}{d^2}$$
Now $$Q_2$$ is halved so to have the same force let the distance be X

$$\Rightarrow \dfrac{K(\dfrac{Q_2}{2})Q_1}{X^2}= \dfrac{K Q_1 Q_2}{d^2}$$

$$\Rightarrow \dfrac{1}{2X^2}= \dfrac{1}{d^2}$$

$$\Rightarrow \dfrac{1}{\sqrt2X}= \dfrac{1}{d}$$

$$\Rightarrow X= \dfrac{d}{\sqrt2}$$
Question 9
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Two unlike charges separated by a distance of $$1m$$ attract each other with a force of $$0.108N$$. If the charges are in the ratio $$1:3$$, the weak charge is :

A
$$2\mu C$$

B
$$4\mu C$$

C
$$6\mu C$$

D
$$5\mu C$$

Solution

Let the charges be $$q $$ and $$3q $$
given force $$= 0.108N$$
$$\Rightarrow 0.108= \dfrac{K(q )(3q )}{(1)^2}$$
$$\Rightarrow 0.108 = 9\times 10^9\times 3q ^2$$

$$\Rightarrow q ^2= \dfrac{108\times 10^{-3}}{9\times 3\times 10^9}$$

$$\Rightarrow q ^2= 4\times 10^{-12}$$
$$\Rightarrow q = 2\times 10^{-6}$$
$$\therefore q = 2 \mu C$$
Question 10
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Two charges $$2C$$ and $$6C$$ are separated by a finite distance. If a charge of $$-4C$$ is added to each of them, The initial force of $$12 \times10^{3}N$$ will change to :

A
$$4\times10^{3}N$$ repulsion

B
$$4\times10^{2}N$$ repulsion

C
$$6\times10^{3}N$$ attraction

D
$$4\times10^{3}N$$ attraction

Solution

Given initial force $$=12\times 10^3$$
$$\Rightarrow \dfrac{k(2)(6)}{x^2}=12\times 10^3$$
when a charge of $$- 4c$$ is added to each of them
$$F_{new}=\dfrac{k(-2)(2)}{x^2}$$
(Putting value of $$x^2$$ from above equation)

$$=\dfrac{k(-2)(2)}{k(2)(6)}\times{12}\times 10^3$$

$$F_{new}=-4\times 10^3N$$ (negative sign implies attraction)

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Electric Charges and Fields Test - 30

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Electric Charges and Fields Test - 30

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Question 1

$$6, 4$$

$$3, 7$$

$$1, 9$$

$$5, 5$$

Solution

Question 2

$$10$$

$$20$$

$$10\sqrt{2}$$

$$2\sqrt{29}$$

Solution

Question 3

1 : 1 : 2

1 : 2 : 2

2 : 2 : 1

2 : 1 : 1

Solution

Question 4

$$92.16\times10^{-3}N $$

$$92.16\times10^{-4}N $$

$$92.16\times10^{-5}N $$

$$92.16\times10^{-2}N $$

Solution

Question 5

$$25$$

$$225$$

$$625$$

$$1250$$

Solution

Question 6

$$0.36\times10^{5}N $$

$$3.6\times10^{5}N $$

$$1.8\times10^{4}dyne $$

$$1.8\times10^{4}N $$

Solution

Question 7

$$2\times10^{9}\ C $$

$$2\times10^{-9}\ C $$

$$\dfrac{2}{3}\times10^{9}\ C $$

$$\dfrac{2}{3}\times10^{-9}\ C $$

Solution

Question 8

$$d$$

$$\dfrac{d}{2}$$

$$\dfrac{d}{\sqrt{2}}$$

$$2d$$

Solution

Question 9

$$2\mu C$$

$$4\mu C$$

$$6\mu C$$

$$5\mu C$$

Solution

Question 10

$$4\times10^{3}N$$ repulsion

$$4\times10^{2}N$$ repulsion

$$6\times10^{3}N$$ attraction

$$4\times10^{3}N$$ attraction

Solution

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