Electric Charges and Fields Test

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Electric Charges and Fields Test - 46

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Question 1
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Two possitively charged spheres of masses $$m_1$$, and $$m_2$$, are suspended from a common point at the ceiling by identical insulating mass-less strings of length I. Charges on the two spheres are $$q_1$$ and $$q_2$$, respectively. At equilibrium both strings make the same angle $$\theta$$ with the vertical. Then :

A
$$q_1m_1=q_2m_2$$

B
$$m_1=m_2$$

C
$$m_1=m_2\sin \theta$$

D
$$q_2m_1=q_1m_2$$

Solution

For equilibrium of $$m_1$$
$$T_1\cos\theta=m_1g ;\
T_1\sin\theta=F$$
$$\tan\theta=\frac{F}{m_1g}$$

For equilibrium of $$m_2$$
$$T_2\cos\theta=m_2g ;\
T_2\sin\theta=F$$
$$\tan\theta=\frac{F}{m_2g}$$
Thus comparing 2 equations $$m_1=m_2$$
Question 2
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The charges on two spheres are $$+7\mu C$$ and $$-5\mu C$$ respectively. They experience a force $$F$$. If each of them is given an additional charge of $$-2\mu C$$, then the new force of attraction will be :

A
$$F$$

B
$$\dfrac { F }{ 2 } $$

C
$$\dfrac { F }{ \sqrt { 3 } } $$

D
$$2F$$

Solution

Given, $${ q }_{ 1 }=+7\mu C=+7\times { 10 }^{ -6 }C$$
$${ q }_{ 2 }=-5\mu C=-5\times { 10 }^{ -6 }C$$
New force $${ F }^{ \prime }=$$?
We know that, $$F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cdot \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$$
$$ F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { \left( +7\times { 10 }^{ -6 } \right) \left( -5\times { 10 }^{ -6 } \right) }{ { r }^{ 2 } }$$
$$ =-\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { 35\times { 10 }^{ -12 } }{ { r }^{ 2 } } N$$
$${ F }^{ \prime }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { \left( +5\times { 10 }^{ -6 } \right) \left( -7\times { 10 }^{ -6 } \right) }{ { r }^{ 2 } }$$
$$ =-\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { 35\times { 10 }^{ -12 } }{ { r }^{ 2 } } N$$
$$\Rightarrow { F }^{ \prime }=F$$
Question 3
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Distance between the two point charges in increased by $$20$$%. Force of interaction between the charges.

A
Increases by $$10$$%

B
Decrease by $$20$$%

C
Decrease by $$17$$%

D
Decrease by $$31$$%

Solution

$$\textbf{Step 1 - Initial electrostatic force}$$
Let the initial distance between charges $$q_{1}$$ and $$q_{2}$$ be $$r$$

$$\therefore$$ Force acting between charge $$(F) = \dfrac {k\cdot q_{1} \cdot q_{2}}{r^{2}}$$ $$....(1)$$

$$\textbf{Step 2-Final electrostatic force}$$
Now, the distance between charges is :-
$$r^{\prime}=r+\dfrac{20}{100}r=r+0.2r$$
$$r' = 1.2 r$$
$$\therefore$$ Force acting between them $$(F') = \dfrac {k\cdot q_{1}q_{2}}{(r')^{2}}$$

$$\Rightarrow F' = \dfrac {k\cdot q_{1} \cdot q_{2}}{(1.2r)^{2}} = \dfrac {k\cdot q_{1} \cdot q_{2}}{1.44r^{2}}$$

$$\Rightarrow F' = \dfrac {F}{1.44}$$ $$....(2)$$

$$\textbf{Step 2 - Calculation of Percentage decrease :}$$
Percentage decrease in force :
$$= \dfrac {F - F'}{F} \times 100$$

$$= \dfrac {F - (F/1.44)}{F} \times 100$$

$$= \dfrac {0.44}{1.44} \times 100$$

$$= 31\%$$

Therefore, force of interaction decreases by $$31\%$$.
Question 4
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Two sphere of electric charges $$+2$$ nC and $$-8$$ nC are placed at a distance '$$d$$' apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

A
$$\dfrac { d }{ 2 } $$

B
$$d$$

C
$$\dfrac { 3d }{ 4 } $$

D
$$\dfrac { 4d }{ 3 } $$

Solution

Initially $$2nC$$ and $$-8 nC$$ are separated by a distance $$d$$.
Electrostatic force acting between them initially $$F = \dfrac{kq_1 q_2}{d^2}$$
$$\therefore$$ $$F = \dfrac{(2)(8) k}{d^2} = \dfrac{16 k}{d^2}$$
Now the two charges are allowed to touch each other, they share equal charges among themselves.
Total charge of the system $$Q = 2-8 = -6 nC$$
Thus new charge on each sphere $$q_1' = q_2'= -3 nC$$
Let the new distance between them be $$d'$$.
Thus force acting between them $$F' = \dfrac{k q_1' q_2'}{(d')^2}$$
$$\therefore$$ $$F' = \dfrac{k(-3)(-3)}{(d')^2} = \dfrac{9 k}{(d')^2}$$
But $$F' = F$$
$$\therefore$$ $$ \dfrac{9 k}{(d')^2} = $$ $$ \dfrac{16 k}{d^2} $$
Or $$(d')^2 = \dfrac{9}{16}d^2$$
$$\implies$$ $$d' = \dfrac{3d}{4}$$
Question 5
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A girl brings a positively charged rod near a thin neutral stream of water flowing from a tap. She observes, that the water stream, bends towards her. Instead, if she is to bring a negativity charged rod near to the stream, it will :

A
bend in the same direction

B
bend in the opposite direction

C
not bend at all

D
bend in the opposite direction above and below the rod

Solution

Water is dipolar in nature. Hence when water enters an electric field, the molecules rearrange themselves.
The case 1 in the figure, where a positively charged rod is brought near the water,
the negative charge moves towards the rod and positive charge moves away from the rod.
Hence, the attractive force is greater than the repulsive force and as a result, the water stream bends towards the rod.

To the contrast, in case 2, when a negatively charged rod is brought near the water, the molecules rearrange such that positive charge moves towards the rod and the negative charge moves away from it as shown in the figure.
In this case also, the attractive forces dominate the repulsive forces, and hence, the stream still bends towards the rod.
Question 6
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A charge of $$0.8$$ coulomb is divided into two charges $${Q}_{1}$$ and $${Q}_{2}$$. These are kept at a separation of $$30cm$$. The force on $${Q}_{1}$$ is maximum when :

A
$${ Q }_{ 1 }={ Q }_{ 2 }=0.4C$$

B
$${ Q }_{ 1 }\approx 0.8C,{ Q }_{ 2 }$$ negligible

C
$${Q}_{1}$$ negligible, $${ Q }_{ 2 }\approx 0.8C$$

D
$${ Q }_{ 1 }=0.2C,{ Q }_{ 2 }=0.6C$$

Solution

Let the two parts be $$(0.8-q)$$ and $$q$$
$$F=\cfrac { k(0.8-q)q }{ { r }^{ 2 } } $$
Putting $$\cfrac { dF }{ dQ } =0$$ we get
$$q=0.4C$$
Question 7
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Two copper balls each weighing 10 g are kept in air 10 cm apart. If one electron from every $$\displaystyle { 10 }^{ 6 }$$ atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is 63.5) :

A
$$\displaystyle 2.0\times { 10 }^{ 10 }N$$

B
$$\displaystyle 2.0\times { 10 }^{ 4 }N$$

C
$$\displaystyle 2.0\times { 10 }^{ 8 }N$$

D
$$\displaystyle 2.0\times { 10 }^{ 6 }N$$

Solution

$$\textbf{Given: }$$ Mass of balls, $$m =10g$$
Distance between the balls, $$ r= 10cm = 0.1m$$

$$\textbf{Step 1: No. of electrons transferred from one ball to another }$$
As number of atoms/molecules in the Mass-number grams of a element is equal to Avogadro number,
Number of atoms in 63.5 g of copper = $$6.022\times10^{23}$$
$$\Rightarrow$$Number of atoms in 1 g of copper = $$\dfrac{6.022\times10^{23}}{63.5}$$
$$\Rightarrow$$Number of atoms in 10g of copper = $$\dfrac{10}{63.5}\times6.022\times10^{23}=0.948\times10^{23}$$

For every $$10^6$$ atoms, one electron is transferred.
$$\therefore$$ Number of electrons transferred from 1 ball to another = $$(n)=0.948\times10^{23}/10^6=0.948\times10^{17}$$

$$\textbf{Step 2: Charge acquired by the balls :}$$

Charge acquired by balls is, $$q=ne=1.6\times10^{-19}\times0.948\times10^{17}=1.52\times10^{-2}C$$

$$\textbf{Step 3: Electrostatic force between the balls :}$$

So electrostatic force between the balls is,
$$F=\dfrac{q^2}{4\pi\epsilon_0r^2}=9\times10^9\times\dfrac{(1.52\times10^{-2})^2}{0.1^2}\approx 2\times10^8N$$

Hence, Option C is correct.
Question 8
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The force between two point charges placed in a material medium of dielectric constant $$\varepsilon_r$$ is $$F$$. If the material is removed, then the force between them becomes:

A
$$\varepsilon_rF$$

B
$$\varepsilon F$$

C
$$\displaystyle\frac{F}{\varepsilon_r}$$

D
$$\displaystyle\frac{\varepsilon}{F}$$

E
$$\displaystyle\varepsilon_0F$$

Solution

$$\textbf{Step 1:Force between point charges in medium}$$
Dielectric constant of medium $$=\varepsilon_r$$
Force in the medium $$=F_m$$
The electric force between point charges in the presence of medium is given by
$$F_{m}=\displaystyle\frac{F_{air}}{\varepsilon_r}$$

$$\textbf{Step 2:Force when medium is removed}$$
When medium is removed then force between charges increased by $$\varepsilon_r$$.
$$F_{air}=F_m\times \varepsilon_r=F\cdot \varepsilon_r$$.
Therefore, If the material is removed, then the force between them becomes $$\varepsilon_rF$$
Question 9
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Two spherical conductors A and B having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor C having same radius as that of A but uncharged is bought in contact with B and then with A. then new force of repulsion between B and A will be :

A
$$F$$

B
$$\dfrac{3F}{4}$$

C
$$\dfrac{F}{8}$$

D
$$\dfrac{3F}{8}$$

Solution

As we know that $$F\propto \dfrac{q^2}{d^2}$$
When B is in contact with C; charge on B and C
$$\dfrac{q+0}{2}=\dfrac{q}{2}$$
When A is in contact with C; charge on A and C
$$\dfrac{q+\dfrac{q}{2}}{2}=\dfrac{3q}{4}$$
$$\therefore $$
Charge on $$B=\dfrac{q}{2}$$Charge on $$A=\dfrac{3q}{4}$$
Force between B and A
$$F_1\propto \dfrac{\dfrac{q}{2}\cdot \dfrac{3q}{2}}{d^2}$$ ....(ii)
From Eqs. (i) and (ii),
$$\dfrac{F_1}{F}=\dfrac{3q^2/8d^2}{q^2/d^2}=\dfrac{3}{8}$$
$$F_1=\dfrac{3}{8}F$$
Question 10
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If electrical force between two charges is $$200 N$$ and we increase $$10$$% charge on one of the charges and decrease $$10$$% charge on the other, then electrical force between them for the same distance becomes :

A
$$200 N$$

B
$$202 N$$

C
$$198 N$$

D
$$199 N$$

Solution

$$\textbf{Hint: Use Coulomb's law}$$

$$\textbf{Step1: Find expression of force in initial case}$$
Let two charges are $${ q }_{ 1 }$$ and $${ q }_{ 2 }$$ and $$r$$ is the distance between them, then electrical force
By using Coulomb's law
$$F=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cdot \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } =200 N$$ ....(i)

$$\textbf{Step2: Find magnitude of force in final case}$$
If $${ q }_{ 1 }$$ is increased by $$10$$%, then
$${ q }_{ 1 }^{ \prime }=(1+\dfrac{10}{100})q_1=\dfrac { 110 }{ 100 } { q }_{ 1 }$$
and $${ q }_{ 2 }$$ is decreased by $$10$$%, then
$${ q }_{ 2 }^{ \prime }=(1-\dfrac{10}{100})q_2=\dfrac { 90 }{ 100 } { q }_{ 2 }$$
Then, electrical force between them,
$${ F }^{ \prime }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \dfrac { { q }_{ 1 }^{ \prime }{ q }_{ 2 }^{ \prime } }{ { r }^{ 2 } }$$
$${ F }^{ \prime }=\dfrac { \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \dfrac { 110 }{ 100 } { q }_{ 1 }\times \dfrac { 90 }{ 100 } { q }_{ 2 } }{ { r }^{ 2 } }$$ .....(ii)
From equations (i) and (ii), we get
$${ F }^{ \prime }=200\times \dfrac { 99 }{ 100 } \Rightarrow { F }^{ \prime }=198N$$
$$\textbf{Hence option C correct}$$

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Electric Charges and Fields Test - 46

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Electric Charges and Fields Test - 46

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Question 1

$$q_1m_1=q_2m_2$$

$$m_1=m_2$$

$$m_1=m_2\sin \theta$$

$$q_2m_1=q_1m_2$$

Solution

Question 2

$$F$$

$$\dfrac { F }{ 2 } $$

$$\dfrac { F }{ \sqrt { 3 } } $$

$$2F$$

Solution

Question 3

Increases by $$10$$%

Decrease by $$20$$%

Decrease by $$17$$%

Decrease by $$31$$%

Solution

Question 4

$$\dfrac { d }{ 2 } $$

$$d$$

$$\dfrac { 3d }{ 4 } $$

$$\dfrac { 4d }{ 3 } $$

Solution

Question 5

bend in the same direction

bend in the opposite direction

not bend at all

bend in the opposite direction above and below the rod

Solution

Question 6

$${ Q }_{ 1 }={ Q }_{ 2 }=0.4C$$

$${ Q }_{ 1 }\approx 0.8C,{ Q }_{ 2 }$$ negligible

$${Q}_{1}$$ negligible, $${ Q }_{ 2 }\approx 0.8C$$

$${ Q }_{ 1 }=0.2C,{ Q }_{ 2 }=0.6C$$

Solution

Question 7

$$\displaystyle 2.0\times { 10 }^{ 10 }N$$

$$\displaystyle 2.0\times { 10 }^{ 4 }N$$

$$\displaystyle 2.0\times { 10 }^{ 8 }N$$

$$\displaystyle 2.0\times { 10 }^{ 6 }N$$

Solution

Question 8

$$\varepsilon_rF$$

$$\varepsilon F$$

$$\displaystyle\frac{F}{\varepsilon_r}$$

$$\displaystyle\frac{\varepsilon}{F}$$

$$\displaystyle\varepsilon_0F$$

Solution

Question 9

$$F$$

$$\dfrac{3F}{4}$$

$$\dfrac{F}{8}$$

$$\dfrac{3F}{8}$$

Solution

Question 10

$$200 N$$

$$202 N$$

$$198 N$$

$$199 N$$

Solution

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