Electric Charges and Fields Test

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Electric Charges and Fields Test - 22

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Question 1
1 / -0

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre

A
Increases as r increases for $r < R$ and for $r > R$

B
Zero as r increases for $r < R$ , decreases as r increases for $r > R$

C
Zero as r increases for $r < R$ , increases as r increases for $r > R$

D
Decreases as r increases for $r < R$ and for $r > R$

Solution

Charge Q will be distributed over the surface of hollow metal sphere.
(i) For r < R (inside)
By Gauss law, $\oint \bar{E_{in}} . \overline{dS} = \dfrac{q_{en}}{\varepsilon_0}= 0$
$\Rightarrow E_{in} = 0$ $(\because q_{en} = 0)$
(ii) For r > R (out side)
$\oint \bar{E}_0 . \overline{dS} = \dfrac{q_{en}}{\varepsilon_0}$
Here, $q_{en} = Q (\because q_{en} = Q)$
$\therefore E_0 4 \pi r^2 = \dfrac{Q}{\varepsilon _0}$
$\therefore E_0 \propto \dfrac{1}{r^2}$
Question 2
1 / -0

Electric field outside a long wire carrying charge q is proportional to :

A
$\dfrac { 1 }{ r }$

B
$\dfrac { 1 }{ { r }^{ 2 } }$

C
$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/5 } } }$

D
$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/2 } } }$

Solution

Using Gauss Law across the cylindrical Gaussian surface,
$\displaystyle\int E.dS = \dfrac{q_{enc}}{\epsilon_0}$
$\therefore E\times 2\pi r \times l = \dfrac{\lambda\times l}{\epsilon_0}$
$\therefore E = \dfrac{\lambda}{2\pi r\epsilon_0}$
Here $\lambda$ is the linear charge density on the wire.
Question 3
1 / -0

An electric line of force in the $xy-plane$ is given by equation $x^{2} + y^{2} = 1$ . A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $xy - plane$ .

A
Not move at all

B
Will move along straight line

C
Will move along the circular line of force

D
Information is insufficient to draw any conclusion

Solution

Charge will move along the circular line of force because $x^{2} + y^{2} = 1$ is the equation of circle in $xy - plane$ .
Question 4
1 / -0

Two identical charged spheres suspended from a common distance $d(d < < 1)$ apart because of their mutual repulsion. The charged begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v$ . Then $v$ varies as a function of the distance $x$ between the spheres, as

A
$v\propto x^{\dfrac {1}{2}}$

B
$v\propto x$

C
$v\propto x^{-\dfrac {1}{2}}$

D
$v\propto x^{-1}$

Solution

From figure $\tan \theta = \dfrac {F_{e}}{mg} =\theta$
$\dfrac {kq^{2}}{x^{2}mg} = \dfrac {x}{2l}$
or $x^{3}\propto q^{2} .... (1)$
or $x^{3/2} \propto q ..... (2)$
Differentiating eq. $(1)$ w.r.t time
$3x^{2} \dfrac {dx}{dt}\propto 2q \dfrac {dq}{dt}$ but $\dfrac {dq}{dt}$ is constant
so $x^{2}(v) \propto q$ Replace $q$ from eq. $(2)$
$x^{2}(v) \propto x^{3/2}$ or $v\propto x^{-1/2}$ .
Question 5
1 / -0

Two unlike charges attract each other with a force of 10N. If the distance between them is doubled, the force between them is :

A
$40N$

B
$20N$

C
$5N$

D
$2.5N$

Solution

We know $F=\dfrac {1}{4\pi \varepsilon_0}\dfrac {q_1q_2}{r^2}=10 N$

if distance is doubled.
$F_{new}=\dfrac {1}{4\pi \varepsilon_0}\cdot \dfrac {q_1q_2}{(2r)^2}=\dfrac {10}{4}$

$=2\cdot 5 N$
Question 6
1 / -0

Number of electric lines of force emanating from $1C$ of positive charge in vacuum is :

A
$8.85\times10 ^{-12}$

B
$9\times10 ^{9}$

C
$1/4\pi \times 9\times 10^{9}$

D
$1.13\times10 ^{11}$

Solution

$\phi$ from $1C$ of charge is $= \dfrac{Q_{enclosed}}{\epsilon_o}$
$= \dfrac{1}{8.8501\times 10^{-12}}$

$= 1.129\times 10^{11}$
Question 7
1 / -0

A force ‘F’ is acting between two charges in air. If the space between them be completely filled with a medium $K=4$ , the force will be :

A
$F$

B
$4F$

C
$\dfrac{F}{4}$

D
$2F$

Solution

By Coulomb's law, the force between two charges $(q_1, q_2)$ in air is given by $F=\dfrac{q_1q_2}{4\pi\epsilon_0 r^2}$
where $r=$ distance between two charges.
Now the force between two charges $(q_1, q_2)$ in medium with dielectric constant $K=4$ becomes, $F'=\dfrac{q_1q_2}{4\pi K\epsilon_0 r^2}$
So, $F'=\dfrac{F}{K}=\dfrac{F}{4}$
Question 8
1 / -0

An electron revolves around the nucleus of hydrogen atom in a circle of radius $5\times 10^{-11}\ m$ . The intensity of electric field at a point in the orbit of the electron is:

A
$5.76\times 10^{11}N/C$

B
$9.216\times 10^{-8}N/C$

C
$0$

D
$4N/C$

Solution

We know hydrogen atom consist of one proton and one electron
$R=5\times 10^{-11} m$
We know attractive force $=\dfrac {k q_1(q_2)}{r^2}=\dfrac {9\times 10^9\times (1\cdot 6\times 10^{-19})^2}{(5\times 10^{-11})^2}$
We know $F = E q$
$\therefore E=\dfrac {F}{q}$
$\therefore E=\dfrac {9\times 10^9\times (1\cdot 6\times 10^{-19})}{5\times 10^{-11})^2}=5\cdot 76\times 10^{11} N/C.$
Question 9
1 / -0

A charge $q_1$ exerts a force of $45N$ on a charge of $q_{2}=10^{-5} C$ located at a point $0.2m$ from $q_{1}$ . The magnitude of $q_{1}$ is

A
$4\times 10^{-8} C$

B
$2\times 10^{-5} C$

C
$3\times 10^{-6} C$

D
$5\times 10^{-8} C$

Solution

As we know electrostatic force between two charges $F=\dfrac {k q_1 q_2}{r^2}$

$\Rightarrow 45=\dfrac {(9\times 10^9)(q_1)(10^{-5})}{(0\cdot 2)^2}$

$\Rightarrow q_1=\dfrac {45\times 0\cdot 2\times 0\cdot 2}{9\times 10^9\times 10^{-5}}$

$\Rightarrow q_1=2\times 10^{-5}C$
Question 10
1 / -0

When we remove polyester or woollen clothes in dark, we can see a spark and hear a crackling sound. Which of the following is responsible for it?

A
Static electricity.

B
Current electricity.

C
Reflection of light.

D
Refraction of light.

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Electric Charges and Fields Test - 22

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Electric Charges and Fields Test - 22

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Question 1

Increases as r increases for r<Rr < Rr<R and for r>R r > Rr>R

Zero as r increases for r<R r < Rr<R, decreases as r increases for r>Rr > R r>R

Zero as r increases for r<Rr < Rr<R, increases as r increases for r>Rr > Rr>R

Decreases as r increases for r<R r < Rr<R and for r>Rr > R r>R

Solution

Question 2

1r\dfrac { 1 }{ r } r1​

1r2\dfrac { 1 }{ { r }^{ 2 } } r21​

1r13/5\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/5 } } } r3/51​1​

1r13/2\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/2 } } } r3/21​1​

Solution

Question 3

Not move at all

Will move along straight line

Will move along the circular line of force

Information is insufficient to draw any conclusion

Solution

Question 4

v∝x12v\propto x^{\dfrac {1}{2}}v∝x21​

v∝xv\propto xv∝x

v∝x−12v\propto x^{-\dfrac {1}{2}}v∝x−21​

v∝x−1v\propto x^{-1}v∝x−1

Solution

Question 5

40N40N40N

20N20N20N

5N5N5N

2.5N2.5N2.5N

Solution

Question 6

8.85×10−128.85\times10 ^{-12}8.85×10−12

9×1099\times10 ^{9}9×109

1/4π×9×1091/4\pi \times 9\times 10^{9}1/4π×9×109

1.13×10111.13\times10 ^{11}1.13×1011

Solution

Question 7

FFF

4F4F4F

F4\dfrac{F}{4}4F​

2F2F2F

Solution

Question 8

5.76×1011N/C5.76\times 10^{11}N/C5.76×1011N/C

9.216×10−8N/C9.216\times 10^{-8}N/C9.216×10−8N/C

000

4N/C4N/C4N/C

Solution

Question 9

4×10−8C4\times 10^{-8} C4×10−8C

2×10−5C2\times 10^{-5} C2×10−5C

3×10−6C3\times 10^{-6} C3×10−6C

5×10−8C5\times 10^{-8} C5×10−8C

Solution

Question 10

Static electricity.

Current electricity.

Reflection of light.

Refraction of light.

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Increases as r increases for $r < R$ and for $r > R$

Zero as r increases for $r < R$ , decreases as r increases for $r > R$

Zero as r increases for $r < R$ , increases as r increases for $r > R$

Decreases as r increases for $r < R$ and for $r > R$

$\dfrac { 1 }{ r }$

$\dfrac { 1 }{ { r }^{ 2 } }$

$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/5 } } }$

$\dfrac { 1 }{ { r }^{ \frac { 1 }{ 3/2 } } }$

$v\propto x^{\dfrac {1}{2}}$

$v\propto x$

$v\propto x^{-\dfrac {1}{2}}$

$v\propto x^{-1}$

$40N$

$20N$

$5N$

$2.5N$

$8.85\times10 ^{-12}$

$9\times10 ^{9}$

$1/4\pi \times 9\times 10^{9}$

$1.13\times10 ^{11}$

$F$

$4F$

$\dfrac{F}{4}$

$2F$

$5.76\times 10^{11}N/C$

$9.216\times 10^{-8}N/C$

$0$

$4N/C$

$4\times 10^{-8} C$

$2\times 10^{-5} C$

$3\times 10^{-6} C$

$5\times 10^{-8} C$