Electric Charges and Fields Test

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Electric Charges and Fields Test - 55

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Question 1
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A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from each other in air?

A
$$\dfrac{q}{2}$$ and $$\dfrac{q}{2}$$

B
$$\dfrac{q}{4}$$ and $$\dfrac{3q}{4}$$

C
$$\dfrac{q}{3}$$ and $$\dfrac{2q}{3}$$

D
$$\dfrac{q}{5}$$ and $$\dfrac{4q}{5}$$

Solution

Suppose Q on one sphere &
q - Q on other sphere
total charge is still $$a+q-Q=q$$
placed at distance r (constant)
then force = $$\dfrac{1}{4\pi c_{10}}\dfrac{Q(q- Q)}{r^{2}}$$
$$F= K$$ $$Q(q-Q)$$ $$ik= \dfrac{1}{4\pi c_{10}r^{2}}$$
$$\dfrac{dF}{dQ}= \dfrac{kd}{dQ}\left \langle Qq-Q^{2} \right \rangle= k \left \langle q-2Q \right \rangle$$
For maximum $$\dfrac{dF}{dQ}=0$$
$$\Rightarrow q-2Q=0\Rightarrow Q= q/2$$
so $$F_{max}= \dfrac{1}{4\pi c_{10}}\dfrac{\dfrac{q}{2}.\dfrac{q}{2}}{r^{2}}$$
i.e. For charge values $$\dfrac{q}{2}$$ & $$\dfrac{q}{2}$$
Question 2
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A charge $$+q$$ having mass $$m$$ is released from rest in a uniform electric field $$E$$ momentum acquired by the charge after time $$t$$ is:

A
$$\frac{{{q^2}{E^2}{t^2}}}{{2m}}$$

B
$$\dfrac{q Et}{m}$$

C
$$q Et$$

D
None of these.

Solution

initial velocity $$u=0,$$
acceleration $$a=qE/m,$$
time $$t=t,$$
final velocity $$v=?$$
by $$v=u+at,$$ we get
$$v = 0 + \left( {\frac{{qE}}{m}} \right)t$$
$$or\,v = \frac{{qEt}}{m}$$
so the final kinetic energy is
$$K\left( t \right) = \frac{1}{2}m{\left( {\frac{{qEt}}{m}} \right)^2}$$
$$ \Rightarrow K\left( t \right) = \frac{{{q^2}{E^2}{t^2}}}{{2m}}$$
Hence,
option $$(A)$$ is correct answer.
Question 3
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A plane surface of area $$10\ cm^{2}$$ is placed in a uniform electric field of $$20\ N/C$$ such that the angle between the surface and the electric field is $$30^{o}$$. The electric flux over the surface is:

A
$$0.0173\ Vm$$

B
$$0.0236\ Vm$$

C
$$0.01\ Vm$$

D
$$0.0346\ Vm$$

Solution

Given,
Electric field, $$E=20\,N{{C}^{-1}}$$
Area, $$A=10\times {{10}^{-4}}\,{{m}^{2}}$$ A
Electric Flux $$\phi =E.A\,\cos \theta =20\times 10\times {{10}^{-4}}\cos {{60}^{o}}=0.01\,Vm$$
Question 4
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A wire of linear charge density $$\lambda$$ passes through a cuboid of length $$\ell$$, breadth $$b$$ and height $$h\ (\ell>b>h)$$ in such a manner that flux through the cuboid is maximum. The position of the wire is now changed, so that the flux through the cuboid is minimum. The ratio of maximum flux to minimum flux will be :

A
$$\dfrac{\sqrt{l^{2}+b^{2}}}{h}$$

B
$$\dfrac{\sqrt{l^{2}+b^{2}+h^{2}}}{h}$$

C
$$\dfrac{h}{\sqrt{l^{2}+b^{2}}}$$

D
$$\dfrac{l}{\sqrt{l^{2}+b^{2}+h^{2}}}$$

Solution

Formula,

$$Q=\dfrac{q}{\epsilon _0}$$

maximum flux occurs diagonally,

$$Q_{max}=\dfrac{\sqrt{L^2+B^2+H^2}}{\epsilon _0}$$

minimum flux occurs when height equals length,

$$Q_{min}=\dfrac{H}{ \epsilon _0}$$

the ratio of maximum by minimum flux is given by,

$$\dfrac{Q_{max}}{Q_{min}}=\dfrac{\dfrac{\sqrt{L^2+B^2+H^2}}{\epsilon _0}}{\dfrac{H}{ \epsilon _0}}$$

$$\therefore \dfrac{Q_{max}}{Q_{min}}=\dfrac{\sqrt{L^2+B^2+H^2}}{H}$$
Question 5
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Two points charges $$+\ 9e$$ and $$+e$$ are kept $$16\ cm$$ apart to each other. Where a third charge $$q$$ is placed between them so that the system is in the equilibrium state-

A
$$24\ cm$$ form $$+9e$$

B
$$12\ cm$$ form $$+9e$$

C
$$24\ cm$$ form $$+e$$

D
$$12\ cm$$ form $$+e$$

Solution

For equilibrium state
$$\dfrac { 9\times { q }_{ c } }{ { x }^{ 2 } } =\dfrac { q\times e }{ { \left( 16-x \right) }^{ 2 } } $$
or, $$\dfrac { 9 }{ { x }^{ 2 } } =\dfrac { 1 }{ { \left( 16-x \right) }^{ 2 } } $$
or, $$\dfrac { 3 }{ x } =\dfrac { 1 }{ 16-x } $$
or, $$3\left( 16-x \right) =x$$
or, $$48-3x=x$$
or, $$4x=48$$
or, $$x=12$$ (Ans)
Now, from $$x$$ the distance will be $$(16-12)cm=4cm$$.
Question 6
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Force of attraction between two point electric charges placed at a distance d in a medium is F. What distance apart should these be kept in the same medium, so that force between them becomes $$\frac{F}{3}$$?

A
$$2 \sqrt 3 d$$

B
$$3d$$

C
$$9d$$

D
$$\sqrt 3 d$$

Solution

$$\textbf{Step 1: Force between the charges}$$
Let the charges be $$q_{1}$$ and $${q}_{2}$$

When kept $$'d'$$ distance apart, the columbic force between them is given by,
$$\displaystyle F=k\dfrac{{q}_{1}{q}_{2}}{{d}^{2}} $$ $$....(1)$$

When charges get separated by $${d}^{\prime }$$, force reduces to $$F'= {F}/{3}$$
$$\Rightarrow \dfrac{F}{3}= k\dfrac{{q}_{1}{q}_{2}}{{d'}^{2}}$$ $$....(2)$$

$$\textbf{Step 2: Solving equations }$$
Dividing Equation $$(1)$$ by $$(2) \Rightarrow \ \ 3 = \cfrac{d'^2}{d^2}$$
$$\Rightarrow\ \ \displaystyle 3{d}^{2}={d}^{\prime 2}\ \ \Rightarrow\ \ {d}^{\prime }=\sqrt{3}d$$

Hence, option $$(D)$$ is correct.
Question 7
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An electron enters an electric field with its velocity in the direction of the electric lines of field then:

A
the path of the electron will be a circle

B
the path of the electron will be a parabola

C
the velocity of the electron will decrease just after entry

D
the velocity of the electron will increase just after entry

Solution

When an electron enters an electric field with its velocity in the direction of the electric lines of field, then the velocity of the electron will decrease just after entry due to the lines of force of the electric field.
Question 8
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There are two charges $$+1\mu$$ and $$+5\mu$$ . The ratio of the forces acting on them will be

A
$$1:5$$

B
$$1:1$$

C
$$5:1$$

D
$$1:25$$

Solution
Question 9
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In the fig. force on charge at $$A$$ in the direction normal to $$BC$$ will be:

A
$$-\dfrac {kq}{a^{2}}$$

B
$$-\dfrac {kq}{2a^{2}}$$

C
$$\dfrac {kq}{2a^{2}}$$

D
zero

Solution

Hence force experienced by the charge $$b$$ at $$A$$ in the direction normal to $$BC$$ is zero.
Question 10
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Charges $$\theta_1$$ and $$\theta_2$$ lie inside and out side respectively for a closed surface s. Let E be the field at a point on S and $$\phi$$ be flux of E over S

A
If only $$\theta_1$$ changes, both E & $$\phi$$ charges

B
If only $$\theta_2$$ changes, E will change but $$\phi$$ will not change

C
If $$\theta_1=0, \theta_2\neq 0$$ then $$E \neq 0$$ but $$\phi = 0$$

D
If $$\theta_1=0, \theta_2\neq 0$$ then $$E = 0$$ but $$\phi \neq 0$$

Solution

Charges $${ \theta }_{ 1 }$$ and $${ \theta }_{ 2 }$$ lie inside and out side respectively.
For a closed surface $$=S$$
Let $$E=$$ the field at a point on $$S$$ and $$\varphi $$ be flux of $$E$$ over $$S$$.
In this context if only $${ \theta }_{ 1 }$$ changes, both $$E$$ and $$\varphi $$ charges and if any $${ \theta }_{ 2 }=$$ changes, $$E$$ will change but $$\varphi $$ will not change.
If $${ \theta }_{ 1 }=0$$ and $${ \theta }_{ 2 }\neq 0$$
$$E\neq 0$$ but $$\varphi =0$$
This point is to be noted.