Electric Charges and Fields Test

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Electric Charges and Fields Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

The electric field at a point $$2$$cm from an infinite line charge of linear charge density $$10^{-7}$$ $$Cm^{-1}$$ is?

A
$$4.5\times 10^4NC^{-1}$$

B
$$9\times 10^4NC^{-1}$$

C
$$9\times 10^2NC^{-1}$$

D
$$18\times 10^4NC^{-1}$$

Solution

$$\textbf{Step 1: Electric field due to infinitely long charged wire}$$
Linear charge density of wire, $$\lambda= 10^{-7}Cm^{-1}$$
Electric field at a distance $$r$$ from infinitely long charged wire is given by,
$$E = \dfrac{\lambda }{2\pi\epsilon_o r}$$ $$......(1)$$

$$\textbf{Step 2: Substituting the values in equation (1)}$$
$$ \ E =\dfrac{10^{-7}}{2\pi \times 8.85\times 10^{-12}\times 0.02} =9\times 10^4 \ NC^{-1} $$

Hence, Option (B) is correct.
Question 2
1 / -0

The total solid angle subtended by the sphere is

A
$$0.25\pi$$

B
$$0.5\pi$$

C
$$4\pi$$

D
$$2\pi$$

E
$$1\pi$$

Solution

Solid angle is given by area subtended divided by $$r^2$$.

$$\Omega=\dfrac{\Delta S}{r^2}$$

Now, area subtended by a sphere is the whole surface area of sphere=$$4\pi r^2$$

$$\implies \Omega=\dfrac{4\pi r^2}{r^2}$$

$$\implies \Omega=4\pi$$

Answer-(E)
Question 3
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A rod of length l having charge q uniformly distributed moves towards right with constant speed v. At $$t=0$$, it enters in an imaginary cube of edge $$I/2$$, sketch variation of electric flux passing through the cube with respect to time.

A

B

C

D

Solution

$$t=0$$ to $$t=\displaystyle\frac{I}{2v}$$
At any time t, length of rod inside cube
$$=vt$$
Charge inside cube $$=\displaystyle\frac{q}{I}vt$$
$$\therefore \phi =\displaystyle\frac{q_{In}}{\varepsilon_0}=\frac{qv}{\varepsilon_0I}t, \phi \propto t$$
Graph will be straight line of slope $$qv/\varepsilon_0I$$
$$t=0, \phi =0$$
$$t=\displaystyle\frac{I}{2v}, \phi =\frac{q}{2\varepsilon_0}$$
$$t=I/2v$$ to $$t=I/v$$, half of rod inside cube, $$\phi =\displaystyle\frac{q}{2\varepsilon_0}$$
$$t=\displaystyle\frac{I}{v}$$ to $$t=\displaystyle\frac{3I}{2v}$$
Charge inside the cube,
$$q_{In}=q-\frac{q}{I}\left(vt -\frac{I}{2}\right)$$
$$=\displaystyle\frac{3q}{2}-\frac{qv}{I}$$t
$$\phi =\frac{\displaystyle \left(\frac{3q}{2}-\frac{qv}{I}t\right)}{\varepsilon_0}$$
$$=\displaystyle\frac{3q}{2\varepsilon_0}-\frac{qv}{\varepsilon_0I}t$$
The graph will be straight line of negative slope and positive intercept.
$$t=\displaystyle\frac{I}{v}$$, $$\phi=\displaystyle\frac{q}{2\varepsilon_0}$$
$$t=\displaystyle\frac{3I}{2v}$$, $$\phi =0$$.
Question 4
1 / -0

Which of the following curves shown cannot possibly represent electrostatic field lines?

A

B

C

D

Solution

Electrostatic field lines start or end only at $${90}^{o}$$ to the surface of the conductor. So, curve (b) cannot represent electrostatic field lines are closer.
Question 5
1 / -0

Two small spheres are charged positively, the combined charge being $$5.0 \times 10^{-5} C$$. If each sphere is repelled from in other by a force of 1.0 N when the spheres are 2.0 metre apart, how is the total charge distributed between the spheres?

A
$$4.83 \times 10^{-5} C, 1.17 \times 10^{95} C$$

B
$$5.83 \times 10^{-5} C, 2.17 \times 10^{95} C$$

C
$$3.83 \times 10^{-5} C, 2.17 \times 10^{95} C$$

D
$$3.85 \times 10^{-5} C, 1.15 \times 10^{-5} C$$

Solution
Question 6
1 / -0

Under the action of a given coulombic force the acceleration of an electron is $$\times { 10 }^{ 22 }m\quad { s }^{ -2 }$$. Then the magnitude of the acceleration of a proton under the action of same force is nearly

A
$$1.6\times { 10 }^{ -19 }m\quad { s }^{ -2 }$$

B
$$9.1\times { 10 }^{ 31 }m\quad { s }^{ -2 }$$

C
$$1.5\times { 10 }^{ 19 }m\quad { s }^{ -2 }\quad $$

D
$$0.6\times { 10 }^{ 27 }m\quad { s }^{ -2 }$$

Solution

The acceleration due to given coulombic force $$F$$ is
$$a=\cfrac { F }{ m } or\quad a\propto \cfrac { 1 }{ m } ......(i)$$
$$\therefore \cfrac { { a }_{ p } }{ { a }_{ e } } =\cfrac { { m }_{ e } }{ { m }_{ p } } $$, where $${m}_{e}$$ and $${m}_{p}$$ are masses of electron and proton respectively
$${ a }_{ p }=\cfrac { { a }_{ e }{ m }_{ e } }{ { m }_{ p } } =\cfrac { \left( 2.5\times { 10 }^{ 22 }m{ s }^{ -2 } \right) \left( 9.1\times { 10 }^{ -31 }kg \right) }{ 1.67\times { 10 }^{ -27 }kg } =13.6\times { 10 }^{ 18 }m{ s }^{ -2 }\simeq 1.5\times { 10 }^{ 19 }m{ s }^{ -2 }$$
Question 7
1 / -0

A proton is kept at rest. A positively charged particle is released from rest at a distance $$d$$ in its field. Consider two experiments; one in which the charged particle is a proton and in another a positron. In the same time $$t$$, the work done on the two moving charged particles is

A
Same as the same force law is involved in the two experiments

B
Less for the case of a positron, as the positron moves away more rapidly and the force on it weakens

C
More for the case of a positron, as the positron moves away a larger distance

D
Same as the work done by charged particle on the stationary proton

Solution

Force between two protons $$=$$ force between a proton and a positron. As positron is much lighter than proton, it moves away a larger distance compared to proton. As work done $$= force\times distance$$, therefore in the same time $$t$$, work done in case of positron is more than that in case of proton.
Question 8
1 / -0

A field of $$100V{ m }^{ -1 }$$ is directed at $${30}^{o}$$ to positive x-axis. Find $$\left( { V }_{ A }-{ V }_{ B } \right) $$ if $$OA=2m$$ and $$OB=4m$$.

A
$$100\left( \sqrt { 3 } -2 \right) V$$

B
$$-100\left( 2+\sqrt { 3 } \right) V\quad $$

C
$$100\left( 2-\sqrt { 3 } \right) V$$

D
$$200\left( 2+\sqrt { 3 } \right) V$$

Solution

A field of $$=100v/m$$ directed $$={ 30 }^{ 0 }$$
$${ V }_{ A }-{ V }_{ B }=?$$
$$OA=2m$$
$$OB=4m$$
Exaplanation
$$\overline { E } =\left( 100\cos{ 30 }^{ 0 }\hat { i } +100\sin{ 30 }^{ 0 }\hat { j } \right) $$
$$\overline { F } =50\sqrt { 3 } \hat { i } +50\hat { j } $$
$$A=\left( 2m,0 \right) $$
$$B=\left( 0,4m \right) $$
$${ V }_{ A }-{ V }_{ B }=-\int _{ B }^{ A }{ \overline { E } .d\overline { r } } $$
$$=\int _{ \left( -2,0 \right) }^{ \left( 0,4 \right) }{ 50\sqrt { 3 } dx } +50dy$$
$$={ \left[ 50\sqrt { 3 } x+50y \right] }_{ \left( 2,0 \right) }^{ \left( 0,4 \right) }$$
$$=-100\left( 2-\sqrt { 3 } \right) $$
Question 9
1 / -0

Two identical small bodies each of mass $$m$$ and charge $$q$$ are suspended from two strings each of length $$l$$ from a fixed point. This whole system is taken into an orbiting artificial satellites, then find the tension in strings

A
$$\dfrac{{K{q^2}}}{{{l^2}}} + 2mg$$

B
$$\dfrac{{K{q^2}}}{{4{l^2}}} + 2mg$$

C
$$\dfrac{{K{q^2}}}{{{l^2}}}$$

D
$$\dfrac{{K{q^2}}}{{4{l^2}}}$$

Solution

Mass $$=m$$
charge $$=q$$
length $$=l$$
now, come to the point, the two bodies are kept in a satellite. Hence, there is no gravity to act on them. So, the only force that acts on the bodies is the coulombs repulsion force. This force repels the two charges until there is maximum separation on between them.
Now, the maximum separation between the two will be the length of the strings.
So, $$r=2l$$
Hence, the tension is the string is due to the force of repulsion.
$$T={ F }_{ c }=\dfrac { Kq }{ { r }^{ 2 } } $$
$$T=\dfrac { Kq }{ { \left( 2l \right) }^{ 2 } } =\dfrac { Kq }{ { 4l }^{ 2 } } $$
Question 10
1 / -0

Two point charges separated by a distance $$d$$ repel each with a force of $$9N.$$ If the separation between them becomes $$3d,$$ the force repulsion will be

A
1 N

B
3 N

C
6 N

D
27 N

Solution

Let the two charges be $${ Q }_{ 1 }$$ & $${ Q }_{ 2 }$$ and according to coulom's force,
$$\vec { F } =\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ { R }^{ 2 } } $$
where, ATQ, $$\vec { F } =9N$$
$$R=d$$
$$\Rightarrow \quad 9=\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ { d }^{ 2 } } $$
$${ 9d }^{ 2 }=K{ Q }_{ 1 }{ Q }_{ 2 }\quad \longrightarrow (1)$$
Now, in $${ 2 }^{ nd }$$ case when $$d\rightarrow 3d$$, force is given by,
$${ \vec { F } }^{ 1 }=\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ { \left( 3d \right) }^{ 2 } } =\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ 9{ d }^{ 2 } } \quad \longrightarrow (2)$$
Substituting (2) into (1) will result to
$${ \vec { F } }^{ 1 }=\dfrac { K{ Q }_{ 1 }{ Q }_{ 2 } }{ K{ Q }_{ 1 }{ Q }_{ 2 } } =1N$$

$$\therefore $$ Option (A) is the correct option.