Electric Charges and Fields Test

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Electric Charges and Fields Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Two identical tiny metal balls carry charges of $$+3nC$$ and $$-12nC$$. They are $$3cm$$ apart. The balls are now touched together and then separated to $$3cm$$. The force on them is :

A
$$20.25\times 10^{-4}N$$

B
$$2.025\times 10^{-3}N$$

C
$$2.025\times 10^{-4}N$$

D
$$2025N$$

Solution

When the balls are touched charge flows from high potential to low potential till potential on both becomes equal $$(V= \dfrac{Kq}{R})$$
As balls are identical, finally they both have equal charges i.e, $$\dfrac {total \ charge}{2}$$ $$=\dfrac{+3-12}{2}=4.5nc$$

force on them (repulsion)$$ F= \dfrac{K(4.5nC)(4.5nC)}{(3\times 10^{-2})^2}$$

$$= \dfrac{9\times 10^9\times 4.5\times 10^{-9}\times 4.5\times10^{-9}}{9\times 10^{-4}}$$

$$= 4.5 \times 4.5 \times 10^{-5}$$
$$= 20.25\times 10^{-5}N $$
$$= 2.025\times 10^{-4}N$$
Question 2
1 / -0

A charge of $$1\mu C$$ is divided into two parts such that their charges are in the ratio of $$2 : 3$$. These two charges are kept at a distance 1m apart in vaccum. Then, the electric force between them (in newton) is :

A
$$0.216$$

B
$$0.00216$$

C
$$0.0216$$

D
$$2.16$$

Solution

Let the charges be

$$2q $$ and $$3q $$

$$2q +3q = 1\mu C$$

$$5q = 1\mu C$$

$$q = \dfrac{1}{5}\mu C$$

$$\therefore charges\ are\ \dfrac{2}{5}\mu C \ and\ \dfrac{3}{5}\mu C$$

Force $$= \dfrac{Kq _1 q _2}{r^2}$$

$$= \dfrac{9\times 10^9\times }{1^2}\dfrac{2}{5}\times 10^{-6}\times \dfrac{3}{5}\times 10^{-6}$$

$$= \dfrac{9\times 6\times 10^{-3}}{25}$$

$$= 2.16\times 10^{-3}$$

$$= 0.00216$$
Question 3
1 / -0

Two charges $$Q_{1}$$ and $$Q_{2}$$ are placed in vacuum at a distance d and the force acting between them is $$5$$ units. If a medium of dielectric constant $$2$$ is introduced around them, the net force now will be :

A
$$10$$

B
$$2.5$$

C
$$5$$

D
infinite

Solution

when dielectric is added $$\varepsilon _0$$ becomes $$k\varepsilon _0$$
where $$k$$ is dielectric constant
we known $$F_i=\dfrac{1}{4\pi \ \varepsilon_0}\dfrac{Q_1Q_2}{r^2}$$
when dielectric is added $$f_f=\dfrac{1}{4\pi \ k\varepsilon _0}\dfrac{Q_1Q_2}{r^2}$$
$$\therefore F_f =\dfrac{5}{2}$$ $$=2\cdot 5\ units$$
Question 4
1 / -0

If the force between two charged objects is to be left unchanged even though the charge on one of the object is halved keeping the other same, the original distance (d) of separation should be changed to :

A
$$2d$$

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

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

D
$$d$$

Solution

initial force $$=\dfrac{k\ q_1q _2}{d^2}$$
Now $$q _1$$ is halved, let final distance be x for same force to act
$$\Rightarrow \dfrac{k(\dfrac{q_1}{2})(q _2)}{x^2}=\dfrac{k\ q_1q_2}{d^2}$$

$$\Rightarrow \dfrac{1}{\sqrt{2}x}=\dfrac{1}{d}$$

$$\Rightarrow x=\dfrac{d}{\sqrt{2}}$$
Question 5
1 / -0

Two identical metallic spheres A and B carry charges $$+Q$$ and $$-2Q$$ respectively. The force between them is 'F' newton, when they are separated by a distance 'd' in air. The spheres are allowed to touch each other and are moved back to their initial position. The force between them now is:

A
$$2F$$

B
$$\dfrac{F}{8}$$

C
$$\dfrac{F}{4}$$

D
$$8F$$

Solution

intial force $$F_i=\dfrac{k(Q)((2Q)}{d^2}$$
$$F_i=\dfrac{2k\ Q^2}{d^2}$$
Now when spheres are touched, charge distribution changes to make potential equal on both spheres
$$\therefore Q_f = \dfrac{Q}{2} $$ (as both the spheres are identical)

$$F_f=\dfrac{k\left ( \dfrac{Q}{2} \right )\left ( \dfrac{Q}{2} \right )}{d^2}$$

$$F_f=\dfrac{k\ Q^2}{4\ d^2}$$

$$F_f=\dfrac{F}{8}$$
Question 6
1 / -0

A particle that carries a charge $$-q$$ is placed at rest in uniform electric field $$10\ N/C$$. It experiences a force and moves in a certain time t, it is observed to acquire a velocity $$10\vec{i}-10\vec{j}$$ m/s. The given electric field intersects a surface of area $$A$$ $$m^{2}$$in the X -Z plane. Electric flux through surface is:

A
$$5\sqrt{2}A\ Nm^{2}/C$$

B
$$5A\ Nm^{2}/C$$

C
$$\sqrt{2}A\ Nm^{2}/C$$

D
$$2\sqrt{5}A\ Nm^{2}/C$$

Solution

$$v=10\hat{i}-10\hat{j}$$

As velocity is in x and y components
Electric field also has two components

Let $$E=Ex\hat{i}+Ey\hat{j}$$

Now force on $$(-q)
=E(-q)$$

Force $$F=-Exq\hat{i}+-Eyq\hat{j}$$

Acceleration $$a =(-\dfrac{Exq}{m})\hat{i} +(-\dfrac{EyV}{m})\hat{i}$$

Now $$v_x=10$$ and $$v_y=-10$$

we know $$v =u+at \ \ (given \ u=0 )$$

$$\therefore V=at$$

$$\Rightarrow v_x=(-\dfrac{Exq}{m})t --(1) \ \ and\ \ v_y=(\dfrac{-Eyq}{m})t --(2)$$

$$Given |E|=10\ N/C$$

Now from 1 and 2 we get,

$$-1=\dfrac{-Exq}{m}\dfrac{m}{Eyq}$$

$$\Rightarrow Ex=Ey $$

$$10=\sqrt{Ex^{2}+Ey^{2}}$$

$$\Rightarrow Ex=5\sqrt{2}=Ey$$

Now than $$\sigma = \int E.dA$$

As the area is in X-Z plane

Area =$$A$$$$\hat{i}$$

$$\therefore Ea=Ey.A$$

$$\phi=5\sqrt{2}A$$
Question 7
1 / -0

Two point charges $$+4C$$ and $$+6C$$ repel each other with a force F. If a charge $$-5C$$ is given to each of these charges, the force becomes:

A
repulsive force of $$\dfrac{F}{24}$$

B
attractive force of $$\dfrac{F}{24}$$

C
repulsive force $$24F$$

D
attractive force $$24F$$

Solution

$$\Rightarrow Given\ F = \dfrac{K(4)(6)}{r^2}$$
Initially as both charges have similar sign there exist repulsion force when $$-5C$$ is given the charges become opposite and attract each other.
$$F_{new}= \dfrac{K(-1)(1)}{r^2}$$

$$\dfrac{F_{new}}{F}= \dfrac{\dfrac{-K}{r^2}}{\dfrac{K\times 24}{r^2}}$$

$$F_{new}= -\dfrac{F}{24}$$
(negative sign here implies the attractive nature of force)
Question 8
1 / -0

Two point charges of $$2C$$ and $$-6C$$ attract each other with a force of $$12N$$. A charge -$$2C$$ is added to $$- 6C$$ charge now, then the force between them is :

A
$$12N$$

B
$$16N$$

C
$$32N$$

D
$$24N$$

Solution

Let distance between the charges be X
$$\Rightarrow \dfrac{K(2)(6)}{X^2}= 12N$$

When $$-2C$$ charge is added to $$-6C$$ the charges are.
$$F= \dfrac{K(2)(8)}{X^2}$$

Put value of $$X^2$$:

$$F = \dfrac{K(2)(8)}{K(2)(6)}\times 12$$

$$F= 16N$$
Question 9
1 / -0

A cube is arranged such that its length, breadth, height are along X,Y and Z directions. One of its corners is situated at the origin. Length of each side of the cube is $$25\ cm$$ . The components of electric field are $$E_{x}=400\sqrt{2}\ N/C,\ E_{y}=0$$ and $$E_{Z}=0$$ respectively. The flux coming out of the cube at one end, whose plane is perpendicular to X axis, will be:

A
$$25\sqrt{2}\ Nm^{2}/C$$

B
$$5\sqrt{2}\ Nm^{2}/C$$

C
$$250\sqrt{2}\ Nm^{2}/C$$

D
$$25\ Nm^{2}/C$$

Solution

Flux coming out on one side is $$\vec E.d \vec a$$

So flux coming out of the cube through the surface normal to X-axis is given by:

$$ \phi =E_x \times (25\times 25\times 10^{-4})$$ $$ =\dfrac{400 \sqrt{2}\times 25\times 25}{10000}$$$$ =25\sqrt{2}$$ $$Nm^2/C$$
Question 10
1 / -0

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

A
1 : 1

B
1 : 2

C
4 : 1

D
2 : 1

Solution

In the given problem, if F is force of repulsion between two charges. Then two charges $$25\mu C$$ and $$50\mu C$$ exerts F force on each other. Hence ratio of forces on them is 1 : 1.

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Electric Charges and Fields Test - 31

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Question 1

$$20.25\times 10^{-4}N$$

$$2.025\times 10^{-3}N$$

$$2.025\times 10^{-4}N$$

$$2025N$$

Solution

Question 2

$$0.216$$

$$0.00216$$

$$0.0216$$

$$2.16$$

Solution

Question 3

$$10$$

$$2.5$$

$$5$$

infinite

Solution

Question 4

$$2d$$

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

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

$$d$$

Solution

Question 5

$$2F$$

$$\dfrac{F}{8}$$

$$\dfrac{F}{4}$$

$$8F$$

Solution

Question 6

$$5\sqrt{2}A\ Nm^{2}/C$$

$$5A\ Nm^{2}/C$$

$$\sqrt{2}A\ Nm^{2}/C$$

$$2\sqrt{5}A\ Nm^{2}/C$$

Solution

Question 7

repulsive force of $$\dfrac{F}{24}$$

attractive force of $$\dfrac{F}{24}$$

repulsive force $$24F$$

attractive force $$24F$$

Solution

Question 8

$$12N$$

$$16N$$

$$32N$$

$$24N$$

Solution

Question 9

$$25\sqrt{2}\ Nm^{2}/C$$

$$5\sqrt{2}\ Nm^{2}/C$$

$$250\sqrt{2}\ Nm^{2}/C$$

$$25\ Nm^{2}/C$$

Solution

Question 10

1 : 1

1 : 2

4 : 1

2 : 1

Solution

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