Electric Charges and Fields Test

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Electric Charges and Fields Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The field at a distance $$r$$ from a long straight wire of charge per unit length $$\lambda$$ is:

A
$$k \dfrac {\lambda}{r^{2}}$$

B
$$k \dfrac {\lambda}{r}$$

C
$$k \dfrac {\lambda}{2r}$$

D
$$k \dfrac {2\lambda}{r}$$

Solution

The field at a distance $$=r$$ a long straight wire of charge per unit length $$\lambda $$.
then the field $$V=k\dfrac { 2\lambda }{ r } $$
Question 2
1 / -0

The solid angle subtended by the periphery of an area $$1 cm^2$$ at a point situated symmetrically at a distance of 5 cm from the area is

A
$$2 \times 10^{-2}$$ steradian

B
$$4 \times 10^{-2}$$ steradian

C
$$6 \times 10^{-2}$$ steradian

D
$$8 \times 10^{-2}$$ steradian

Solution

Given data,
$$Area=1cm^{2}$$
$$Distance=5cm$$
$$Solid \ angle \ \Omega=?$$
We know the formula
Solid angle $$\Omega=\dfrac{Area}{(Radial \ distance)^{2}}$$

$$=\dfrac{1cm^{2}}{(5cm^{2})}=\dfrac{1}{25}$$

$$=4\times 10^{-2}steradian$$
The solid angle is for $$3-D$$ figure like spere, cone, etc and plane angle is for plane objects or $$2-D$$ figure like circle,arc etc.
Question 3
1 / -0

The forces between two small charged spheres having charges of $$1\times { 10 }^{ -7 }C$$ and $$2\times { 10 }^{ -7 }C$$ placed $$20cm$$ apart in air is:

A
$$4.5\times { 10 }^{ -2 }N$$

B
$$4.5\times { 10 }^{ -3 }N$$

C
$$5.4\times { 10 }^{ -2 }N$$

D
$$5.4\times { 10 }^{ -3 }N$$

Solution

Here, $${ q }_{ 1 }=1\times { 10 }^{ -7 },{ q }_{ 2 }=2\times { 10 }^{ -7 }\quad ,r=20cm=20\times { 10 }^{ -2 }m$$

As $$\quad F=\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } =\cfrac { 9\times { 10 }^{ 9 }\times 1\times { 10 }^{ -7 }\times 2\times { 10 }^{ -7 } }{ { \left( 20\times { 10 }^{ -2 } \right) }^{ 2 } } =4.5\times { 10 }^{ -3 }N$$
Question 4
1 / -0

Two point charges exert on each other a force $$F$$ when they are placed $$r$$ distant apart in air. When they are placed $$R$$ distance apart in a medium of dielectric constant K, they were exerting the same force. The distance $$R$$ equals

A
$$\dfrac {r}{\sqrt {K}}$$

B
$$\dfrac {r}{K}$$

C
$$rK$$

D
$$r\sqrt {K}$$

Solution

Given two point charges exert a force $$F$$ when placed $$r$$ apart in air. When they are placed $$R$$ apart inn a medium of dielectric constant $$K$$, the force between them is same. We have to find $$R$$.

For this, let $$q_{1}$$ and $$q_{2}$$ be two point charges. When these charges are placed in dielectric medium, then force is $$\dfrac{1}{4\pi \varepsilon }\dfrac{q_{1}q_{2}}{R^{2}}$$
where $$\dfrac{\varepsilon }{\varepsilon _{0}}=K=$$ dielectric constant of the medium.
So, new force $$=\dfrac{1}{4\pi \varepsilon_{0}K}\dfrac{q_{1}q_{2}}{R^{2}}$$
when placed in air, $$F=\dfrac{1}{4\pi \varepsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}$$
$$\varepsilon =$$ permittivity of medium
$$\varepsilon _{0}=$$ permittivity of air

Given that force in both the cases is same; then $$\dfrac{1}{4\pi \varepsilon }\dfrac{q_{1}q_{2}}{r^{2}}=\dfrac{1}{4\pi \varepsilon }\dfrac{q_{1}q_{2}}{KR^{2}}$$
$$r^{2}=KR^{2}$$
So, $$R=\dfrac{r}{\sqrt{K}}$$
Question 5
1 / -0

The constant $$k$$ in Coulomb's law depends upon

A
nature of medium

B
system of units

C
intensity of charge

D
both (a) and (b)

Solution

The value of $$k=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } =8.854\times { 10 }^{ -12 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$$
where $${ \varepsilon }_{ 0 }$$ is permittivity of free space
Question 6
1 / -0

The electrostatic attracting force on a small sphere of charge $$0.2\mu C$$ due to another small sphere of charge $$-0.4\mu C$$ in air is $$0.4N$$. The distance between the two spheres is

A
$$43.2\times { 10 }^{ -6 }m\quad $$

B
$$42.4\times { 10 }^{ -3 }m$$

C
$$18.1\times { 10 }^{ -3 }m$$

D
$$19.2\times { 10 }^{ -6 }m$$

Solution

Here, $${ q }_{ 1 }=0.2\mu C=0.2\times { 10 }^{ -6 }C,\quad { q }_{ 2 }=.-0.4\mu C=-0.4\times { 10 }^{ -6 }C,F=-0.4N$$
As $$F=\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } $$
$${ r }^{ 2 }=\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ 4\pi { \varepsilon }_{ 0 }F } =\cfrac { 0.2\times { 10 }^{ -6 }\times 0.4\times { 10 }^{ -6 }\times 9\times { 10 }^{ 9 } }{ 0.4 } =1.8\times { 10 }^{ -3 }\quad $$
$$\therefore r={ \left( 1.8\times { 10 }^{ -3 } \right) }^{ 1/2 }=0.0424m=42.4\times { 10 }^{ -3 }m$$
Question 7
1 / -0

A change of $$1\ \mu C$$ is divided into two parts such that their charges are in the ratio of 2:3. These two charges are kept at a distance 1 m apart in vaccum. Then, the electric force between them (in Newton) is

A
0.216

B
0.00216

C
0.0216

D
2.16

Solution

Given that,
$$ 1\mu C\quad$$ charge is divided in ratio $$ 2:3$$ and kept at distance.
$$d=1m$$ apart. to find electric force between them.
so, two parts have charges
$$ \cfrac { 2 }{ 5 } \times 1\mu C\quad and\quad \cfrac { 3 }{ 5 } \times 1\mu C\\ ie,{ q }_{ 1 }=\cfrac { 2 }{ 5 } \mu C\quad and\cfrac { 3 }{ 5 } \mu C={ q }_{ 2 }\\ d=1m$$
so; coulomb's force b/w these 2 parts is
$$ f=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \\ =\cfrac { 9\times { 10 }^{ 9 } }{ { \left( 1 \right) }^{ 2 } } \times \cfrac { 2 }{ 5 } \times \cfrac { 3 }{ 5 } \times { 10 }^{ -12 }\\ =2.16\times { 10 }^{ -3 }=0.00216N$$
so the answer is option (b).
Question 8
1 / -0

Two insulated charged metallic spheres $$P$$ and $$Q$$ have their centres separated by a distance of $$60cm$$. The radii of $$P$$ and $$Q$$ are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is $$3.2\times { 10 }^{ -7 }C$$ is:

A
$$5.2\times { 10 }^{ -4 }N$$

B
$$2.56 \times { 10 }^{ -3 }N$$

C
$$1.5\times { 10 }^{ -3 }N$$

D
$$3.5\times { 10 }^{ -4 }N$$

Solution

Here
$${q}_{1}={q}_{2}=3.2\times {10}^{-7}C,r=60cm=0.6m$$
Electrostatic force,
$$F=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } =\cfrac { 9\times { 10 }^{ -3 }{ \left( 3.2\times { 10 }^{ -3 } \right) }^{ 2 } }{ { \left( 0.6 \right) }^{ 2 } } =2.56\times { 10 }^{ -3 }N$$
Question 9
1 / -0

The acceleration for electron and proton due to electrical force of their mutual attraction when they are $$1\mathring {A}$$ apart is

A
$$3.1\times { 10 }^{ 22 }m{ s }^{ -2 },1.3\times { 10 }^{ 19 }m { s }^{ -2 }$$

B
$$3.3\times { 10 }^{ 18 }m { s }^{ -2 },3.2\times { 10 }^{ 16 }m { s }^{ -2 }$$

C
$$2.5\times { 10 }^{ 22 }m { s }^{ -2 },1.4\times { 10 }^{ 19 }m { s }^{ -2 }$$

D
$$2.5\times { 10 }^{ 18 }m { s }^{ -2 },1.3\times { 10 }^{ 16 }m { s }^{ -2 }$$

Solution

$$\textbf{Step 1: Force between electron and proton } $$
Distance between electron and proton, $$r = 1\ \mathring{A}= 10^{-10}m$$

From Coulomb's law, $$F = \dfrac{1}{4 \pi \epsilon_{0}} \dfrac{e^2}{r^2}$$ $$= \dfrac{9 \times 10^{9} \times (1.6 \times 10^{-19}C)^{2}}{(10^{-10}m)^{2}}= 2.3\times 10^{-8}\ N$$

$$\textbf{Step 2: Acceleration of proton and electron}$$
From newton's law, $$F=ma$$
Acceleration of electron $$ = \dfrac{F}{m_{e}}$$$$ = \dfrac{2.3 \times 10^{-8}N}{9 \times 10^{-31}\ kg} $$ $$= 2.5 \times 10^{22}\ m/s^{2} $$
Acceleration of proton $$ = \dfrac{F}{m_{p}}$$$$ = \dfrac{2.3 \times 10^{-8}N}{1.66 \times 10^{-27}\ kg} $$ $$ = 1.4 \times 10^{19}\ m/s^{2}$$

Hence, Option(C) is correct.
Question 10
1 / -0

A charge $$Q$$ is placed at the centre of the line joining two point charges $$+q$$ and $$+q$$ as shown in figure, The system is in equilibrium. The ratio of charges $$Q$$ and $$q$$ is

A
$$4$$

B
$$1/4$$

C
$$-4$$

D
$$-1/4$$

Solution

$$\textbf{Step 1: Equilibrium of central charge Q }$$$$\textbf{[Ref. Fig. 1]}$$
For the system to be in equilibrium net force on each charge should be zero.

By Symmetry, both outer charges $$q$$ will exert equal and opposite force on $$Q$$, which will cancel out. As shown in Figure 1.
Hence Net force on $$Q$$ will always be zero irrespective of value of charges.

$$\textbf{Step 2: Equilibrium of outer charge q }$$$$\textbf{[Ref. Fig. 2]}$$
Net force on charge $$q$$ at the right, due to other two charges should be zero.
$$F_{q} = \dfrac{KQq}{x^2} + \dfrac{Kq^2}{(2x)^2} = 0$$

$$\Rightarrow\ \ \dfrac{KQq}{x^2} = -\dfrac{Kq^2}{4x^2}$$

$$\Rightarrow\ \ \dfrac{Q}{q} = \dfrac{-1}{4}$$

Hence option $$(C)$$ correct.

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Electric Charges and Fields Test - 49

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Electric Charges and Fields Test - 49

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Question 1

$$k \dfrac {\lambda}{r^{2}}$$

$$k \dfrac {\lambda}{r}$$

$$k \dfrac {\lambda}{2r}$$

$$k \dfrac {2\lambda}{r}$$

Solution

Question 2

$$2 \times 10^{-2}$$ steradian

$$4 \times 10^{-2}$$ steradian

$$6 \times 10^{-2}$$ steradian

$$8 \times 10^{-2}$$ steradian

Solution

Question 3

$$4.5\times { 10 }^{ -2 }N$$

$$4.5\times { 10 }^{ -3 }N$$

$$5.4\times { 10 }^{ -2 }N$$

$$5.4\times { 10 }^{ -3 }N$$

Solution

Question 4

$$\dfrac {r}{\sqrt {K}}$$

$$\dfrac {r}{K}$$

$$rK$$

$$r\sqrt {K}$$

Solution

Question 5

nature of medium

system of units

intensity of charge

both (a) and (b)

Solution

Question 6

$$43.2\times { 10 }^{ -6 }m\quad $$

$$42.4\times { 10 }^{ -3 }m$$

$$18.1\times { 10 }^{ -3 }m$$

$$19.2\times { 10 }^{ -6 }m$$

Solution

Question 7

0.216

0.00216

0.0216

2.16

Solution

Question 8

$$5.2\times { 10 }^{ -4 }N$$

$$2.56 \times { 10 }^{ -3 }N$$

$$1.5\times { 10 }^{ -3 }N$$

$$3.5\times { 10 }^{ -4 }N$$

Solution

Question 9

$$3.1\times { 10 }^{ 22 }m{ s }^{ -2 },1.3\times { 10 }^{ 19 }m { s }^{ -2 }$$

$$3.3\times { 10 }^{ 18 }m { s }^{ -2 },3.2\times { 10 }^{ 16 }m { s }^{ -2 }$$

$$2.5\times { 10 }^{ 22 }m { s }^{ -2 },1.4\times { 10 }^{ 19 }m { s }^{ -2 }$$

$$2.5\times { 10 }^{ 18 }m { s }^{ -2 },1.3\times { 10 }^{ 16 }m { s }^{ -2 }$$

Solution

Question 10

$$4$$

$$1/4$$

$$-4$$

$$-1/4$$

Solution

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