Electric Charges and Fields Test

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Electric Charges and Fields Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

When two charges are equal q each, force they exert on each other is $$F$$. When one of the charges is doubled, the $$2$$q charge exerts a force $$2F$$ on charge q. The force exerted by q on $$2$$q is :

A
$$F$$

B
$$\displaystyle \dfrac{F}{2}$$

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

D
$$2F$$

Solution

Electrostatic force is mutual. Force exerted on charge $$q$$ due to charge $$2q$$ is equal to the force exerted on $$2q$$ due to charge $$q$$. So, if $$2q$$ charge exerts a force $$2F$$ on charge $$q$$, then the force exerted by $$q$$ on $$2q$$ is also $$2F$$.
Ans:(D)
Question 2
1 / -0

Two charged particles each of mass $$5$$g and charge $$q=2\mu C $$ are suspended as shown in the figure. The system is taken in a satellite. The force between the charges is:

A
$$23 \times 10^{-3} N$$

B
$$2.3 \times 10^{-3} N$$

C
$$0.23 \times 10^{-3} N$$

D
none of these

Solution

Given : $$l = 60$$ cm $$q = 2\times 10^{-6}$$ C
As there is no gravity, thus total distance between the charges $$d =2l+5 = 125$$ cm $$=1.25$$ m
Coulomb force $$F = \dfrac{Kq^2}{d^2} = \dfrac{9\times 10^9\times (2\times 10^{-6})^2}{(1.25)^2} = 23\times 10^{-3}$$ N
Question 3
1 / -0

A charge Q is divided into two parts. The two charges kept at a distance apart have a maximum columbian repulsion. Then the ratio of Q and one of the parts is given by :

A
$$1 : 4$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$4 : 1$$

Solution

Total charge is Q
Now let it is divided into q and (Q-q)
Force of repulsion is $$F=\cfrac{q(Q-q)}{r^2}$$
For F to be maximum $$\cfrac{dF}{dq}=0$$
$$\Rightarrow \cfrac{Q-q}{r^2}-\cfrac{q}{r^2}=0$$
$$\cfrac{Q}{q}=2$$
Question 4
1 / -0

The field at a distance r from a long string of charge per unit length $$\lambda$$ is :

A
$$\displaystyle k \dfrac{\lambda}{r^2}$$

B
$$\displaystyle k \dfrac{\lambda}{r}$$

C
$$\displaystyle k \dfrac{\lambda}{2r}$$

D
$$\displaystyle k \dfrac{2 \lambda}{r}$$

Solution

Consider a surface of a cylinder of radius $$r$$ and length $$l$$ as Gaussian surface. Apply Gauss's law, $$\displaystyle \vec{E}.\vec{ds}=\dfrac{Q_{en}}{\epsilon_0}$$ where $$Q_{en}$$ is the charge enclosed by the Gaussian surface. here $$Q_{en}=\lambda l$$

thus, $$\displaystyle E.(2\pi rl)=\dfrac{\lambda l}{\epsilon_0}$$

$$\displaystyle \therefore E=\dfrac{\lambda }{2\pi \epsilon_0 r}=k\dfrac{2\lambda }{r}$$ where $$\displaystyle k=\dfrac{1 }{4\pi \epsilon_0 r}$$
Question 5
1 / -0

A charge Q is divided into two parts q and Q - q and separated by a distance R. The force of repulsion between them will be maximum when

A
$$\displaystyle q = Q/4$$

B
$$\displaystyle q = Q/2$$

C
$$q = Q$$

D
none of the above

Solution

We have force of repulsion between the two charges,

$$ F_r = \dfrac {1}{4 \pi \varepsilon_0} \dfrac {q (Q - q)}{ R^2}$$

Now

(a)if $$ q = \dfrac {Q}{4}$$, force $$F = \dfrac{1}{4 \pi \varepsilon_0} \dfrac {3Q}{16}$$

(b)if $$q = \dfrac {Q}{2}$$, force $$F = \dfrac {1}{4 \pi \varepsilon_0} \dfrac {Q}{4}$$

(c) if $$ q = Q$$, force $$F = 0$$

Hence the force of repulsion between them will be maximum when $$q = \dfrac {Q}{2}$$.
Question 6
1 / -0

In a certain region of space, there exists a uniform electric field of $$2\times 10^{3} \hat {k}\ Vm^{-1}$$. A rectangular coil of dimension $$10\ cm\times 20\ cm$$ is placed in $$x - y$$ plane. The electric flux through the coil is

A
Zero

B
$$30\ Vm$$

C
$$40\ Vm$$

D
$$50\ Vm$$

Solution

$$\vec{E}=2000\vec{k},\vec{A}= 10 \times 20 \times 10^{-4} \vec{k}$$
$$\phi= \vec{E}\cdot \vec{A}=2000\times 10 \times 20 \times 10^{-4}Vm=40Vm$$
Question 7
1 / -0

In comparison to the electrostatic force between two electrons, the electrostatic force between two proton is :

A
zero

B
smaller

C
same

D
greater

Solution

Since the magnitude of charge of electron and proton is same, $$e$$, the electrostatic force between two proton is same compared to the electrostatic force between two electrons.
Question 8
1 / -0

Directions For Questions

Two point like charges $$Q_1$$ and $$Q_2$$ are positioned at points 1 and 2. The field intensity to the right of the charge $$Q_2$$ on the line that passes through the two charges varies according to a law that is represented schematically in figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x-axis. The distance between the charges is $$l$$.

...view full instructions

The sign of each charge $$Q_1$$ and $$Q_2$$ is:

A
+, -

B
-, +

C
+, +

D
-, -

Solution

Electric field at (2) tends to $$- \infty$$, hence the charge at (2) should be negative. There is a neutral point to the right of charges. This is possible only when the charge at (1) should be positive. Hence, $$Q_1$$ is positive and $$Q_2$$ is negative. At neutral point, $$\vec E_1 = \vec E_2$$
$$\displaystyle \frac{1}{4 \pi \varepsilon_0} \frac{Q}{(a + l)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{a^2} or \frac{Q_1}{Q_2} = \left ( \frac{a + l}{a} \right )^2$$
Electric field at any position on the right of charges is
$$\displaystyle E = E_1 - E_2 = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{(l + x)^2} - \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{x^2}$$
For the maximum value of E,
$$\displaystyle \frac{dE}{dx} = 0 or \frac{d}{dx} \left [ \frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{(l + x)^2} - \frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{x^2} \right ] = 0$$
or $$\displaystyle Q_1 \frac{(-2)}{(1 + x)^3} - Q_2 \frac{(-2)}{x^3} = 0$$
or $$\displaystyle \frac{Q_1}{(l + x)^3} = \frac{Q_2}{x^3} or \left ( \frac{l + x}{x}\right )^3 = \frac{Q_1}{Q_2}$$
or $$\displaystyle \frac{l + x}{x} = \left ( \frac{Q_1}{Q_2} \right )^{1/3} or \frac{l}{x} = \left ( \frac{Q_1}{Q_2} \right )^{1/3} - 1$$
or $$\displaystyle x = \frac{l}{\left ( \frac{Q_1}{Q_2} \right )^{1/3} - 1} $$
Question 9
1 / -0

Two identical conducting small spheres are placed with their centres $$0.3 m$$ apart. One is given a charge of $$12.0 nC$$ and the other a charge of $$-18.0 nC$$. Find the electric force exerted by one sphere on the other?

A
$$4.16 \times 10^{-5} N$$

B
$$2.16 \times 10^{-5} N$$

C
$$6.16 \times 10^{-5} N$$

D
$$7.66 \times 10^{-5} N$$

Solution

Using Coulomb's law, $$F=\dfrac{1}{4\pi \epsilon_0} \dfrac{q_1q_2}{r^2}=(9\times 10^9)\times \dfrac{(12\times 10^{-9})(-18\times 10^{-9})}{(0.3)^2}=-2.16\times 10^{-5} N$$ (negative sign for attractive force)
Question 10
1 / -0

A cylinder of length $$L$$ and radius $$b$$ has its axis coincident with the x-axis. The electric field in this region is $$\vec {E} = 200\hat {i}$$. Find the flux through the left end of cylinder.

A
$$0$$

B
$$200\pi b^{2}$$

C
$$100\pi b^{2}$$

D
$$-200\pi b^{2}$$

Solution

$$\vec E=200\hat{i}$$
Area vector for the left end is $$\vec A=-A\hat{i}$$
$$ \phi=\vec E.\vec A =-EA=-E \pi b^2 =- 200 \pi b^2$$

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Electric Charges and Fields Test - 34

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Electric Charges and Fields Test - 34

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Question 1

$$F$$

$$\displaystyle \dfrac{F}{2}$$

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

$$2F$$

Solution

Question 2

$$23 \times 10^{-3} N$$

$$2.3 \times 10^{-3} N$$

$$0.23 \times 10^{-3} N$$

none of these

Solution

Question 3

$$1 : 4$$

$$1 : 2$$

$$2 : 1$$

$$4 : 1$$

Solution

Question 4

$$\displaystyle k \dfrac{\lambda}{r^2}$$

$$\displaystyle k \dfrac{\lambda}{r}$$

$$\displaystyle k \dfrac{\lambda}{2r}$$

$$\displaystyle k \dfrac{2 \lambda}{r}$$

Solution

Question 5

$$\displaystyle q = Q/4$$

$$\displaystyle q = Q/2$$

$$q = Q$$

none of the above

Solution

Question 6

Zero

$$30\ Vm$$

$$40\ Vm$$

$$50\ Vm$$

Solution

Question 7

zero

smaller

same

greater

Solution

Question 8

+, -

-, +

+, +

-, -

Solution

Question 9

$$4.16 \times 10^{-5} N$$

$$2.16 \times 10^{-5} N$$

$$6.16 \times 10^{-5} N$$

$$7.66 \times 10^{-5} N$$

Solution

Question 10

$$0$$

$$200\pi b^{2}$$

$$100\pi b^{2}$$

$$-200\pi b^{2}$$

Solution

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