Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Figure given below shows a network of four capacitors of capacitance equal to $${ C }_{ 1 }=C$$, $${ C }_{ 2 }=2C$$, $${ C }_{ 3 }=3C$$, $${ C }_{ 4 }=4C$$ connected to a battery. The ratio of charges on $${ C }_{ 2 }$$ and $${ C }_{ 4 }$$ is :

A
$${ 7 }/{ 4 }$$

B
$${ 7 }/{ 22 }$$

C
$${ 3 }/{ 22 }$$

D
$${ 4 }/{ 7 }$$

Solution

The charge flowing through $${ C }_{ 4 }$$ is
$${ q }_{ 4 }={ C }_{ 4 }\times V=4{ C }_{ V }$$
The series combination of $${ C }_{ 1 }$$, $${ C }_{ 2 }$$ and $${ C }_{ 3 }$$
$$\dfrac { 1 }{ { C }^{ \prime } } =\dfrac { 1 }{ C } +\dfrac { 1 }{ 2C } +\dfrac { 1 }{ 3C } $$
$$\dfrac { 1 }{ { C }^{ \prime } } =\dfrac { 6+3+2 }{ 6C } =\dfrac { 11 }{ 6C } \Rightarrow { C }^{ \prime }=\dfrac { 6C }{ 11 } $$
Now, $${ C }^{ \prime }$$ and $${ C }_{ 4 }$$ form parallel combination giving
$${ C }^{ \prime \prime }={ C }^{ \prime }+{ C }_{ 4 }=\dfrac { 6C }{ 11 } +4C=\dfrac { 50C }{ 11 }$$
Net charge, $$ q={ C }^{ \prime \prime }V=\dfrac { 50 }{ 11 } CV$$
Total charge flowing through $${ C }_{ 1 }$$, $${ C }_{ 2 }$$ and $${ C }_{ 3 }$$ will be
$${ q }^{ \prime }=q-{ q }_{ 4 }=\dfrac { 50 }{ 11 } CV-4CV=\dfrac { 6CV }{ 11 }$$
Since, $$ { C }_{ 1 }$$, $${ C }_{ 2 }$$, $${ C }_{ 3 }$$ are in series combination. Hence, charge flowing through these will be same.
Hence, $${ q }_{ 2 }={ q }_{ 1 }={ q }_{ 3 }={ q }^{ \prime }=\dfrac { 6CV }{ 11 }$$
Thus, $$ \dfrac { { q }_{ 2 } }{ { q }_{ 4 } } =\dfrac { { 6CV }/{ 11 } }{ 4CV } =\dfrac { 3 }{ 22 } $$
Question 2
1 / -0

A parallel plate air capacitor has capacity $$'C'$$ farad, potential $$'V'$$ volt and energy $$'E'$$ joule. When the gap between the plates is completely filled with dielectric.

A
Both $$V$$ and $$E$$ increase

B
Both $$V$$ and $$E$$ decrease

C
$$V$$ decreases, $$E$$ increases

D
$$V$$ increases, $$E$$ decreases

Solution

A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely proportional to dielectric $$V'= \dfrac{V}{K}$$ where $$K$$ is dielectric constant.
Question 3
1 / -0

A tin nucleus has charge $$50eV$$. If the proton is $$ 10^{-12} m $$ from the nucleus. Then, the potential at this position will be :

A
$$ 7.2 \times 10^8 V $$

B
$$ 3.6 \times 10^4 V $$

C
$$ 1.44 \times 10^4 V $$

D
$$ 7.2 \times 10^4 V $$

Solution

Charge on the nucleus
Q = 50 eV
$$ = 50 \times 1.6 \times 10^{-19} = 80 \times 10^{-19} $$
Distance of proton from nucleus
$$ r = 10^{-12} m $$
Potential of this position is given by
$$ V = \dfrac {1}{4 \pi l_0 } . \dfrac {Q}{r} $$
$$ = 9 \times 10^9 \times \dfrac {80 \times 10^{-19}}{10^{-12}} $$
$$ = 7.2 \times 10^4 V $$
Question 4
1 / -0

Three identical capacitors are first connected in series and then in parallel. The ratio of effective capacitances in the two cases is

A
$$9 : 1$$

B
$$3 : 1$$

C
$$1 : 3$$

D
$$1 : 9$$

Solution

Let the capacitance of each capacitor be $$C$$.
Effective capacitance in series connection $$\dfrac{1}{C_{s}} = \dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}$$
$$\therefore$$ $$C_s = \dfrac{C}{3}$$
Effective capacitance in parallel connection $$C_p = C+C+C$$
$$\therefore$$ $$C_p = 3C$$
Thus ratio of effective capacitance $$\dfrac{C_s}{C_p} = \dfrac{C/3}{3C}$$
$$\implies$$ $$C_s:C_p = 1:9$$
Question 5
1 / -0

Two parallel plate air capacitors of same capacity $$'C'$$ are connected in series to a battery of emf $$'E'$$. Then one of the capacitors is completely filled with dielectric material of constant $$'K'$$. The change in the effective capacity of the series combination is

A
$$\dfrac {C}{2}\left [\dfrac {K - 1}{K + 1}\right ]$$

B
$$\dfrac {2}{C}\left [\dfrac {K - 1}{K + 1}\right ]$$

C
$$\dfrac {C}{2}\left [\dfrac {K + 1}{K - 1}\right ]$$

D
$$\dfrac {C}{2}\left [\dfrac {K - 1}{K + 1}\right ]^{2}$$

Solution

Initial capacitance $$C_i=\dfrac{1}{C}+\dfrac{1}{C}$$, $$C_i=\dfrac{C}{2}$$
Final $$\dfrac{1}{C_f}=\dfrac{1}{KC}+\dfrac{1}{C}$$
so $$C_f=\dfrac{CK}{K+1}$$
$$C_f-C_i=\dfrac{C}{2}[\dfrac{K-1}{K+1}]$$
Question 6
1 / -0

Each capacitor shown in figure is $$ 2 \mu F . $$ Then the equivalent capacitance between A and B is :

A
$$ 2 \mu F $$

B
$$ 4 \mu F $$

C
$$ 6 \mu F$$

D
$$ 8 \mu F$$

Solution

Equivalent capacitance of upper arms in series
$$ C_1 = \dfrac {2 \times 2}{2 + 2} = 1 \mu F $$
In lower arms $$C_2 = 1 \mu F $$
Now $$ C_1 $$ and $$ C_2 $$ are in parallel
$$ \therefore $$ Equivalent capacitance $$ C = C_1 + C_2 = 2 \mu F $$
Question 7
1 / -0

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is $$3.75\mu F$$. The capacity of each capacitor is

A
$$5\mu F$$

B
$$6\mu F$$

C
$$7\mu F$$

D
$$8\mu F$$

Solution

Let the capacitance be C
when connected parallel, $$C_1=C+C+C=3C$$
when other was connected in series
So $$\dfrac{1}{C_f}=\dfrac{1}{3.75}=\dfrac{1}{3C}+\dfrac{1}{C}$$
$$C=5\mu F$$
Question 8
1 / -0

Capacitance of a capacitor becomes $$ \frac {7}{6} $$ times its original value, if a dielectric slab of thickness $$ t = \frac {2}{3} d $$ is introduced in between the plates. The dielectic constant of the dielectric slab is :

A
$$\frac{14}{11} $$

B
$$\frac{11}{14} $$

C
$$\frac{7}{11} $$

D
$$\frac{11}{7} $$

Solution

The capacitance is given by $$ C_1 = \frac {\epsilon_0 A }{d} $$ ...(i)
If electric slab of constant k of thickness $$ t = \frac {2}{3} d $$ is introduced, then capacitance becomes
$$ C_2 = \frac {\epsilon_0 A}{d - t \left( 1 - \frac {1}{k} \right) } $$
$$ = \frac {\epsilon_0 A }{d - \frac{2}{3} d \left( 1 - \frac {1}{k} \right) } $$ ...(ii)
Eq. (i) becomes
$$ C_2 = \frac {\epsilon_0 A }{d \begin{bmatrix} \left( 1 - \frac {2}{3} \right) + \frac {2}{3k} \end{bmatrix} } $$
$$\Rightarrow C_{ 2 }=\frac { C_{ 1 } }{ \left( \frac { 1 }{ 3 } +\frac { 2 }{ 3k } \right) } $$
$$ \Rightarrow \frac {7}{6} C_1 = \frac {C_1}{\frac {1}{3} + \frac {2}{3k}} $$
$$ \Rightarrow \frac {1}{3} + \frac {2}{3k} = \frac {6}{7} $$
$$ \Rightarrow \frac {2}{3k} = \frac {6}{7} - \frac {1}{3} = \frac {11}{21} $$
or $$ 33k = 21 \times 2 $$
$$ k = \frac {42}{33} = \frac {14}{11} $$
Question 9
1 / -0

$$A \ \ 6 \times 10^{-4}$$ F parallel plate air capacitor is connected to a 500 V battery. When air is replaced by another dielectric material, $$7.5 \times 10^{-4} C$$ charge flows into the capacitor. The value of the dielectric constant of the material is:

A
1.5

B
2.0

C
1.0025

D
3.5

Solution

Let the dielectric constant be $$K$$.
Capacitance of parallel plate air capacitor $$C = 6\times 10^{-4} \ F$$
Voltage of the battery $$V =500$$ volts
Thus charge stored on air capacitor $$Q_a = CV = (7.5\times 10^{-4})(500) = 3000 \times 10^{-4} $$ C
Extra charge flown $$Q = 7.5\times 10^{-4} $$ C
Thus total charge stored on dielectric capacitor $$Q' = (3000+7.5)\times 10^{-4} = 3007.5\times 10^{-4}$$ C
Thus capacitance of parallel plate dielectric capacitor $$C = \dfrac{Q'}{V} = \dfrac{3007.5\times 10^{-4}}{500} =6.015\times 10^{-4} \ F$$
Capacitance of parallel plate dielectric capacitor $$C' = KC$$
$$\therefore$$ $$.015\times 10^{-4} = K(6\times 10^{-4})$$
$$\implies \ K = 1.0025$$
Question 10
1 / -0

The total capacitance of the system of capacitors in figure. between A and B is

A
$$1 \mu F$$

B
$$2 \mu F$$

C
$$3 \mu F$$

D
$$4 \mu F$$

Solution

For series combination :
$$C_{eq} = \dfrac{C_1C_2}{C_1+ C_2}$$

For parallel combination :
$$C_{eq}=C_1 + C_2$$

The outer $$2 \mu F $$ and $$ 2 \mu F $$ are in series, hence their equivalent capacitance will be $$1 \mu F$$.

Now this $$1 \mu F$$ will be parallel to $$1 \mu F$$, hence their equivalent will become $$2 \mu F$$.

Now this $$2 \mu F $$ will be in series with $$2\mu F$$ capacitor whose equivalent will become $$1 \mu F$$.

And this $$1 \mu F$$ will be in parallel with last $$1 \mu F$$, due to which equivalent capacitance will become $$2 \mu F$$ which is the required one.

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Electrostatic Potential and Capacitance Test - 48

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Electrostatic Potential and Capacitance Test - 48

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Question 1

$${ 7 }/{ 4 }$$

$${ 7 }/{ 22 }$$

$${ 3 }/{ 22 }$$

$${ 4 }/{ 7 }$$

Solution

Question 2

Both $$V$$ and $$E$$ increase

Both $$V$$ and $$E$$ decrease

$$V$$ decreases, $$E$$ increases

$$V$$ increases, $$E$$ decreases

Solution

Question 3

$$ 7.2 \times 10^8 V $$

$$ 3.6 \times 10^4 V $$

$$ 1.44 \times 10^4 V $$

$$ 7.2 \times 10^4 V $$

Solution

Question 4

$$9 : 1$$

$$3 : 1$$

$$1 : 3$$

$$1 : 9$$

Solution

Question 5

$$\dfrac {C}{2}\left [\dfrac {K - 1}{K + 1}\right ]$$

$$\dfrac {2}{C}\left [\dfrac {K - 1}{K + 1}\right ]$$

$$\dfrac {C}{2}\left [\dfrac {K + 1}{K - 1}\right ]$$

$$\dfrac {C}{2}\left [\dfrac {K - 1}{K + 1}\right ]^{2}$$

Solution

Question 6

$$ 2 \mu F $$

$$ 4 \mu F $$

$$ 6 \mu F$$

$$ 8 \mu F$$

Solution

Question 7

$$5\mu F$$

$$6\mu F$$

$$7\mu F$$

$$8\mu F$$

Solution

Question 8

$$\frac{14}{11} $$

$$\frac{11}{14} $$

$$\frac{7}{11} $$

$$\frac{11}{7} $$

Solution

Question 9

1.5

2.0

1.0025

3.5

Solution

Question 10

$$1 \mu F$$

$$2 \mu F$$

$$3 \mu F$$

$$4 \mu F$$

Solution

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