Electric Charges and Fields Test

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Electric Charges and Fields Test - 48

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Question 1
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Two small spheres with masses $$M_1$$ and $$M_2$$ hang on weightless, insulating threads with lengths $$L_1$$ and $$L_2$$. The two spheres carry a charge of $$Q_1$$ and $$Q_2$$ respectively. The spheres hang such that they are level with one another and the threads are inclined to the vertical at angles $$\theta_1$$ and $$\theta_2$$. Which of the following conditions is required if $$\theta_1=\theta_2$$.

A
$$M_1=M_2$$

B
$$|Q_1|=|Q_2|$$

C
$$L_1=L_2$$

D
None of these

Solution

$$\theta_1=\theta_2$$, if their masses are the same because the force of repulsion due to charges is the same. The force Mg $$\sin\theta$$ opposes the force of repulsion. Even if $$Q_1$$ and $$Q_2$$ are different, the force of repulsion is the same for both as they are mutual.
Question 2
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Two point charges $$-q$$ and $$+q/2$$ are situated at the origin and the point $$(a,0,0)$$ respectively. The point along the x-axis, where the electric field vanishes is

A
$$x=\sqrt { 2 } a$$

B
$$x=\cfrac { a }{ \sqrt { 2 } } $$

C
$$x=\cfrac { \sqrt { 2 } a }{ \sqrt { 2 } -1 } $$

D
$$x=\cfrac { \sqrt { 2a } }{ \sqrt { 2 } +1 } $$

Solution

The given situation can be shown as given
Suppose the field vanished at a distance $$x$$, we have
$$\cfrac { kq }{ { x }^{ 2 } } =\cfrac { kq/2 }{ { (x-a) }^{ 2 } } $$
$$\Rightarrow 2{ (x-a) }^{ 2 }={ x }^{ 2 }$$
$$\sqrt { 2 } (x-a)=x\quad $$
$$\Rightarrow \sqrt { 2 } x-x=\sqrt { 2 } a$$
$$\Rightarrow (\sqrt { 2 } -1)x=\sqrt { 2 } a$$
$$\quad \Rightarrow x=\cfrac { \sqrt { 2 } a }{ \sqrt { 2 } -1 } $$
Question 3
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The smallest electricforce between two changes placed at a distance of 1 m apart is (approximately)

A
$$10^{-19} \, N$$

B
$$10^{-10} \, N$$

C
$$10^{-28} \, N$$

D
$$10^{-42} \, N$$

Solution

From Coulomb's law, electric force
$$F_e =\frac {1}{4\pi \varepsilon _0}\cdot \frac{q_1q_2}{r^2}$$
For minimum force,
$$q_1=q_2 = 1.6 \times 10^{-19}C$$
$$\therefore F_e = \frac {9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(1)^2}$$
$$f_e = 10^{-28} N$$
Question 4
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If the coulombian force acting between two protons separated by a distance r is F, what would be the force acting between two $$\alpha-$$particles separated by a distance 2r ?

A
$$\dfrac {F}{2}$$

B
$$F$$

C
$$2F$$

D
$$3F$$

Solution

Let the charge of proton is q
Charge of $$\alpha-$$particle
$$ = 2\times charge \, of \, proton$$
$$=2q$$

The force acting between two protons separated by a distance r is
$$F=k \dfrac {q\cdot q}{r^2}=k \dfrac {q^2}{r^2}$$

The force acting between two $$\alpha-$$ patricles separated by a distance 2r is
$$F'=k \dfrac {2q\cdot 2q}{(2r)^2}$$
$$=k \dfrac {4q^2}{4r^2}$$
or $$ = k \dfrac {r^2}{r^2}$$
$$\Rightarrow F'=F$$
Question 5
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The black shapes in the figure are closed surfaces. The electric field lines are in red. For which case the net flux through the surfaces is non-zero?

A
In all cases net flux is non-zero

B
Only (c) and (d)

C
Only (a) and (b)

D
Only (b), (c) and (d)

Solution

As charge is enclosed in a, b, so by Gauss's Law there will be a flux enclosed by the given surface.
Question 6
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Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube .The electric flux through square surface ABCD of the cube is

A
$$\dfrac{q}{24 \in_0}$$

B
$$\dfrac{q}{12 \in_0}$$

C
$$\dfrac{q}{6 \in_0}$$

D
$$\dfrac{q}{8 \in_0}$$

Solution

At each vertex, we have effective charge of $$q_v=q/8$$
So total charge enclosed by the cube is, $$8q_v=q$$

By Gauss law, total electric flux through the cube is, $$\phi=\dfrac{q}{\epsilon_o}$$
By symmetry, flux through each face of the cube would be equal.
So, flux trough ABCD is, $$\dfrac{\phi}{6}=\dfrac{q}{6\epsilon_o}$$

Option C is correct.
Question 7
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Two identical conducting balls A and B have positive charges $$q_1$$ and $$q_2$$ respectively. But $$q_1\neq q_2$$. The balls are brought together so that they touch each other and then kept in their original positions. The force between them is?

A
Less than that before the balls touched

B
greater than that before the balls touched

C
Same as that before the balls touched

D
Zero

Solution

According to coulomb's law, the force of repulsion between them is given by $$F=\displaystyle\frac{q_1q_2}{4\pi \varepsilon_0 r^2}$$
When the charged spheres A and B are brought in contact, each sphere will attain equal charge $$q'$$
$$q'=\displaystyle\frac{q_1+q_2}{2}$$
Now, the force of repulsion between them at the same direction r is
$$F'=\displaystyle\frac{q'\times q'}{4\pi \varepsilon_0 r^2}=\frac{1}{4\pi \varepsilon_0}\frac{\left(\displaystyle\frac{q_1+q_2}{2}\right)\left(\displaystyle\frac{q_1+q_2}{2}\right)}{r^2}$$
$$=\displaystyle\frac{\left(\displaystyle\frac{q_1+q_2}{2}\right)^2}{4\pi \varepsilon_0 r^2}$$
As $$\left(\displaystyle\frac{q_1+q_2}{2}\right)^2 > q_1q_2$$
$$\therefore F' > F$$.
Question 8
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Let the electrostatic field E at distance r from a point charge q not be an inverse square but, instead an inverse cubic, e.g., $$\vec{E}=k\frac{q}{r^3}\hat{r}$$. Here k is a constant. Consider the following two statements.
(i) Flux through a spherical surface enclosing the charge is $$\Phi =q_{enclosed}/\epsilon_0$$
(ii) A charge placed inside uniformly charged shell will experience a force.
Choose the correct option.

A
Only (i) is valid

B
Only (ii) is valid

C
Both (i) and (ii) are invalid

D
Both (i) and (ii) are valid

Solution

Flux enclosed is not constant since $$\int E.ds$$ over a surface of radius r is not constant and is inversely proportional to r. hence (i) is not valid.
Also $$F = Eq$$, initially distance $$r$$ is inverse square in electric field but now distance r is changed to inverse cube in electric field. hence electric field is changed from initial condition, hence a charge will experience a force. hence (ii) is valid.
Question 9
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A certain charge Q is divided into two parts q and $$Q-q$$. How the charge Q and q must be related so that when q and $$(Q-q)$$ is placed at a certain distance apart experience maximum electrostatic repulsion?

A
$$Q=2q$$

B
$$Q=3q$$

C
$$Q=4q$$

D
$$Q=4q+c$$

Solution

The electrostatic force of repulsion between the charge q and $$(Q-q)$$ at separation r is given by
$$\displaystyle F=\frac{1}{4\pi \varepsilon_0}$$ $$\frac{q(Q-q)}{r^2}$$ $$=\frac{1}{4\pi \varepsilon_0}$$ $$\frac{qQ-q^2}{r^2}$$
If F is maximum, then $$\partial F/ \partial q=0$$
i.e., $$\displaystyle\frac{1}{4\pi \varepsilon_0}\cdot \frac{(Q-2q)}{r^2}=0$$
As $$\displaystyle\frac{1}{4\pi \varepsilon_0r^2}$$ is constant, therefore
$$Q-2q=0$$ or $$Q=2q$$
Question 10
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A charge $$Q$$ is situated at the centre of a cube. The electric flux through one of the faces of the cube is

A
$$Q/ \epsilon_{0}$$

B
$$Q/ 2\epsilon_{0}$$

C
$$Q/ 4\epsilon_{0}$$

D
$$Q/ 6\epsilon_{0}$$

Solution

The electric flux through the whole cube:
$$ \phi = \cfrac{Q}{\varepsilon_0}$$
Flux through each face will be one sixth of the total:
$$ \phi = \cfrac{Q}{6\varepsilon_0}$$