Electric Charges and Fields Test

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Electric Charges and Fields Test - 65

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Question 1
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A point charge $$q$$ is revolving in a circle of radius $$' r '$$ around a fixed infinite line charge with positive charge $$\lambda$$ per unit length. Now the point charge is shifted and it revolves in a circle of radius $$ 2 r $$. Then :

A
speed of the point charge $$q$$ remain constant

B
speed of the point charge $$q$$ will be change

C
angular velocity of the point charge $$q$$ about line charge does not change

D
work done by all forces is non-zero

Solution

Electric field at distance $$r$$ from the infinite line charge is $$E=k \dfrac{2 \lambda}{r}$$
The necessary centripetal force to the point charge of mass $$m$$ is provided by electrostatic attraction.
$$\therefore \dfrac{mv^2}{r}=qE=qk \dfrac{2 \lambda}{r}$$ or $$v=\sqrt{\dfrac{2k\lambda q}{m}}$$
The speed of charged particle is independent of orbital radius. Hence the K.E. of point charge in orbits of radius $$r$$ and $$2r$$ is same. As the work done is the change of K.E, so the work done by all forces is zero.
Question 2
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Two similar very small conducting spheres having charges $$40 C$$ and $$ -20 C$$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

A
$$8:1$$

B
$$4:1$$

C
$$1:8$$

D
$$1:1$$

Solution

Initially force between conducting spheres $$F_1=\cfrac{Kq_1q_2}{r^2}=\cfrac{K(40)(-20)}{r^2}$$(i)
After contact total charge$$=40-20=20\mu C$$
Now charge will be equally distributed
so $$q_1=q_2=\dfrac{20}{2}=10\mu C$$
Finally force between spheres$$F_2 =\cfrac{K(10)(10)}{r^2} $$(ii)
Ratio of forces=$$\cfrac{F_1}{F_2}=\cfrac{40\times 20}{10\times 10}=8$$
Question 3
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A long cylindrical wire carries a linear charge density of $$3\times 10^{-8} Cm^{-1}$$. An electron revolve around it in a circular path under the influence of the attractive force. $$KE$$ of the electron is :

A
$$1.44 \times 10^{-7} J$$

B
$$2.88 \times 10^{-17} J$$

C
$$4.32 \times 10^{-17} J$$

D
$$8.64 \times 10^{-17} J$$

Solution

Let $$r$$ be the radius of the circular path.
For long wire the electric field at $$r$$ is $$E=\dfrac{1}{4\pi\epsilon_0}.\dfrac{2\lambda}{r}$$
here, $$\dfrac{mv^2}{r}=eE=\dfrac{e}{4\pi\epsilon_0}.\dfrac{2\lambda}{r}$$
$$\dfrac{1}{2}mv^2=\dfrac{e}{4\pi\epsilon_0}.{\lambda}$$
$$K.E=\dfrac{1}{2}mv^2=\dfrac{1.6\times 10^{-19}}{4\pi\times 8.854\times 10^{-12}}\times 3\times 10^{-8}=4.32 \times 10^{-17} J$$
Question 4
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A dipole is placed in a shell as shown. Find the electric flux emerging out of the shell and in a hypothetical sphere of radius r as shown.

A
$$\displaystyle \frac{2q}{\varepsilon_{0}},0$$

B
$$\displaystyle \frac{q}{\varepsilon_{0}},\, \frac{-q}{\varepsilon_{0}}$$

C
$$\displaystyle \frac{-q}{\varepsilon_{0}},\, \frac{q}{\varepsilon_{0}}$$

D
$$0, \, 0$$

Solution

According to the Gauss's law the electric flux through closed surface is $$\phi=\dfrac{Q_{enclosed}}{\epsilon_0}$$ where $$Q_{enclosed}$$ is the charge inside the closed surface.
Here in both cases, $$Q_{enclosed}=+q-q=0$$
So the electric flux emerging out of the shell and in a hypothetical sphere of radius r are $$(0,0)$$
Question 5
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The charge per unit length for a very long straight wire is $$\lambda$$. The electric field at points near the wire (but outside it) and far from the ends varies with distance r as :

A
r

B
1/r

C
1/r$$^2$$

D
1/r$$^3$$

Solution

This can easily be calculated using Gauss Law as shown in the concept above. which clearly shows that the electric field at points near the wire (but outside it) and far from the ends varies with distance r as $$\frac { 1 }{ r } $$
Question 6
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Suppose that $$1.0\ g$$ of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on earth? (Given that the radius of earth is $$6400\ km$$)

A
$$510\ kN$$

B
$$460\ kN$$

C
$$710\ kN$$

D
$$220 \ kN$$

Solution

Number of electrons or protons in 1.00 g of hydrogen is
$$\displaystyle n = 2 \left [ \frac{1}{2} \times 6.023 \times 10^{23} \right ] = 6.023 \times 10^{23}$$
Magnitude of charge on north or south pole is
$$q = ne = 6.23 \times 10^{23} \times 1.6 \times 10^{-19} = 96368 C$$
$$\displaystyle F = \frac{kq^2}{r^2} = \frac{9 \times 10^9 \times (96368)^2}{(2 \times 6400 \times 10^3)^2} = 5.10 \times 10^5 N$$
$$= 510 kN$$
Question 7
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In the figure shown, A is a fixed charge. Another charge B (of mass m) is given a velocity v perpendicular to line AB. At this moment the radius of curvature of the resultant path of B is-

A
0

B
Infinity

C
$$\dfrac {4\pi \epsilon_0r^2mv^2}{q^2}$$

D
$$r$$

Solution

$$F=\dfrac {K.q^2}{r^2}\Rightarrow \frac {Kq^2}{r^2}=\dfrac {mv^2}{R_C}\Rightarrow R_C=\dfrac {mv^2r^2}{Kq^2}$$
$$R_C=\dfrac {4\pi e_0v^2r^2m}{q^2}$$
Question 8
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A hemisphere of radius r is placed in a uniform electric field of strength E. The electric flux through the hemisphere is :

A
$$2E\pi r^{2}$$

B
$$-E\pi r^{2}$$

C
$$-2E\pi r^{2}$$

D
zero

Solution
Question 9
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Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of $$0.108N$$ when separated by $$0.5m$$. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of $$0.036N$$. The initial charges of the spheres are :

A
$$\pm 5\times { 10 }^{ -6 }C\quad and\quad \mp 15\times { 10 }^{ -6 }C$$

B
$$\pm 1.0\times { 10 }^{ -6 }C\quad and\quad \mp 3.0\times { 10 }^{ -6 }C$$

C
$$\pm 2.0\times { 10 }^{ -6 }C\quad and\quad \mp 6.0\times { 10 }^{ -6 }C$$

D
$$\pm 0.5\times { 10 }^{ -6 }C\quad and\quad \mp 1.5\times { 10 }^{ -6 }C$$

Solution

Let initial charge be Q1 and Q2.
Initial force $$ = F = \displaystyle\frac{K \times Q1 \times Q2}{r^2} = -0.108 $$
$$F = \displaystyle\frac{9 \times 10^9 \times Q1 \times Q2}{0.25} = -0.108 $$

$$ Q1 \cdot Q2 = - 3 \times 10^{-12} $$

Now on connecting them, since they are identical so they will have equal charges.
Final charge $$ = \displaystyle\frac{Q1 + Q2}{2} $$

Final force $$ = F = \displaystyle\frac{9 \times 10^9 \times (Q1+Q2)^2 }{4 \times 0.25} = 0.036 $$
$$ Q1 + Q2 = \pm 2 \times 10^{-6} $$

$$Q2 = \pm 1.0\times { 10 }^{ -6 }C\quad , Q1 = \mp 3.0\times { 10 }^{ -6 }C$$
Question 10
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Two small spheres each of radius 1 mm are kept 10 cm apart. Assuming each proton has a charge +e and each electron has a charge 0.1% less than +e then find the force between the two spheres. Density of copper is 8.9 gcm$$^{-3}$$ and atomic mass number is 63.5.

A
$$1.17 \times 10^12 N$$

B
$$2.34 \times 10^{-12} N$$

C
$$1.17 \times 10^{-12} N$$

D
$$2.34 \times 10^{12} N$$

Solution

mass of 1 mm radius sphere
$$m= 8.9 \displaystyle \times \frac{4}{3} \pi (.1)^3$$
$$= 3.7 \times 10^{-2} g$$
Charge on the sphere$$=\dfrac{m}{A}N_AZ(\dfrac{0.1}{100})\times $$charge of $$e^+$$
$$=\displaystyle

\dfrac{3.7 \times 10^{-2}}{63.5} \times 6.023 \times 10^{23} \times 29

\times \dfrac{.1}{100} \times 1.6 \times 10^{-19}$$
$$= 1.61 C$$
$$\displaystyle F = \dfrac{1.61 \times 1.6 \times 10^{-19} \times 9 \times 10^9}{(0.1)^2}$$
$$= 2.34 \times 10^{12} N$$