Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 49

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Question 1
1 / -0

Effective capacitance between A and B in the figure shown is (all capacitances are in $$\mu F$$).

A
$$\displaystyle\frac{3}{14}\mu F$$

B
$$\displaystyle\frac{14}{3}\mu F$$

C
$$21\mu F$$

D
$$23\mu F$$

Solution

The points C and D will be at same potentials since, $$\displaystyle\frac{3}{6}=\frac{4}{8}$$. Therefore, capacitance of $$2\mu F$$ will be unaffected. So, the equivalent circuit can be shown as the effective capacitance in upper in series, is given by
$$C_1=\displaystyle\frac{3\times 6}{3+6}=\frac{18}{9}=2\mu F$$
The effective capacitance in lower arm in series is given by
$$C_2=\displaystyle\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\mu F$$
Hence, the resultant capacitance in parallel given by
$$C=C_1+C_2=2+\displaystyle\frac{8}{3}=\frac{14}{3}\mu F$$.
Question 2
1 / -0

Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch S. If now the switch S is opened and the space between the plates is filled with a dielectric of relative permittivity $$\displaystyle { \varepsilon }_{ r }$$, then :

A
The potential difference as well as charge on each capacitor goes up by a factor $$\displaystyle { \varepsilon }_{ r }$$

B
The potential difference as well as charge on each capacitor goes down by a factor $$\displaystyle { \varepsilon }_{ r }$$

C
The potential difference across A remains constant and the charge on B remains unchanged

D
The potential difference across B remains constant while the charge on A remains unchanged

Solution

After switch S is opened, as the Capacitor A is connected across the battery, its potential difference is fixed at steady state (i.e., when capacitor is fully charged).

Capacitor B is isolated, so its charge gets fixed. But as we insert the dielectric, its capacitance changes, thus its potential difference also changes.

Option C is correct.
Question 3
1 / -0

Three capacitors connected in series have an effective capacitance of $$4 \mu F$$. If one of the capacitance is removed, the net capacitance of the capacitor increases to $$6 \mu F$$. The removed capacitor has a capacitance of

A
$$2\mu F$$

B
$$4\mu F$$

C
$$10\mu F$$

D
$$12\mu F$$

E
$$24\mu F$$

Solution

Let there are three capacitors with capacitances $$C_1,C_2 , C_3$$ respectively and $$ C_1 $$ is removed.
In first case, $$\dfrac{1}{C_{eq1}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3} $$ ...(1)
In second case, $$\dfrac{1}{C_{eq2}}=\dfrac{1}{C_2}+\dfrac{1}{C_3} $$ ...(2)
From (1) and (2), $$\dfrac{1}{C_{eq1}}=\dfrac{1}{C_1}+\dfrac{1}{C_{eq2}} $$
or $$1/4=1/C_1+1/6$$
or $$C_1=12\, \mu F$$
Question 4
1 / -0

The electrostatic potential energy of two point charges, $$1$$ $$\mu$$C each, placed $$1$$ meter apart in air is?

A
$$9\times 10^3$$J

B
$$9\times 10^9$$J

C
$$9\times 10^{-3}$$J

D
$$9\times 10^{-3}$$eV

Solution

$$\textbf{Step 1: Formula of Potential energy}$$
Electrostatic potential energy is given by $$U$$ between two charges placed at a separtion $$r$$ is given by
$$\Rightarrow\ \ U=\dfrac { k{ q }_{ 1 }{ q }_{ 2 } }{ r} $$

$$\textbf{Step 2: Calculations}$$
Putting values, $$k=9\times { 10 }^{ 9 }\dfrac { N.{ m }^{ 2 } }{ { C }^{ 2 } } ,$$
$${ q }_{ 2 }={ q }_{ 1 }=1\mu C=1\times { 10 }^{ -6 }C,$$
$$r=1m$$

$$\\ \Rightarrow U\\ $$$$ =\dfrac { 9\times { 10 }^{ 9 }\times 1\times { 10 }^{ -6 }\times 1\times { 10 }^{ -6 } }{ 1 } =9\times { 10 }^{ -3 }J\\ $$

Hence, correct option is $$C$$.
Question 5
1 / -0

Capacitance of a capacitor made by a thin metal foil is $$\displaystyle 2\mu F$$. If the foil is folded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, the length of the foil will be :

A
0.34 m

B
1.33 m

C
13.4 m

D
33.9 m

Solution

Here, let the length be $$l$$
width is, $$b=400mm=0.4m$$
thickness, $$t=0.15mm=15\times10^{-5}m$$

As capacitance for parallel plate capacitor is,
$$C=K\epsilon_0A/d=K\epsilon_0(l\times b)/t$$
$$\implies l=Ct/K\epsilon_0b=2\times10^{-6}\times15\times10^{-5}/(2.5\times8.85\times10^{-12}\times0.4)=33.91m$$

Option D is correct.
Question 6
1 / -0

The potential of a large liquid drop when eight liquid drops are combined is $$20\ V$$. Then, the potential of each single drop was :

A
$$10\ V$$

B
$$7.5\ V$$

C
$$5\ V$$

D
$$2.5\ V$$

Solution

Volume of $$8$$ drops $$=$$ Volume of big drop
$$\therefore \left (\dfrac {4}{3}\pi r^{3}\right ) \times 8 = \dfrac {4}{3} \pi R^{3}$$
$$\Rightarrow 2r = R .... (i)$$
According to charge conservation
$$8q = Q .... (ii)$$
We have
$$V \propto \dfrac {1}{r}$$
or $$\dfrac {V_{1}}{V_{2}} = \dfrac {r_{2}}{n}$$
$$\Rightarrow \dfrac {20}{V_{2}} = \dfrac {2r}{r}$$
$$\Rightarrow V_{2} = 10\ V$$
Question 7
1 / -0

If a dielectric introduced between the plates of a parallel plate condenser, then which of the following is possible :

A
Decreases the electric field between the plates

B
Decreases the capacity of the condenser

C
Increases the charge stored in the condenser

D
Increases the capacity of the condenser

Solution

If a dielectric medium of dielectric constant $$K$$ is filled completely between the plates then capacitance increases by $$K$$ times
$$C' = \dfrac {K\epsilon_{0}A}{d}$$
$$\Rightarrow C' = KC$$.
Question 8
1 / -0

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is $$\displaystyle 2\mu F$$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is

A
$$\displaystyle 11.2\mu F$$

B
$$\displaystyle 15.6\mu F$$

C
$$\displaystyle 19.2\mu F$$

D
$$\displaystyle 22.4\mu F$$

Solution

Initial capacitance $$\displaystyle C = 2uF=\frac { { \varepsilon }_{ 0 }{ A }}{ d } $$
New capacitance $$\displaystyle { C }^{ \prime }=\frac { K{ \varepsilon }_{ 0 }A }{ { d }^{ \prime } } =\frac { 2.8\times { \varepsilon }_{ 0 }A }{ (d/2) } =5.6\frac { { \varepsilon }_{ 0 }A }{ d } $$
$$\displaystyle \Rightarrow { C }^{ \prime }= 5.6\times 2=11.2\mu F$$
Question 9
1 / -0

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $$0.4cm$$ is $$2\mu F$$. If the separation is reduced to half and it is filled with a dielectric substance of value $$2.8$$, then the final capacity of the capacitor is

A
$$11.2\mu F$$

B
$$15.6\mu F$$

C
$$19.2\mu F$$

D
$$22.4\mu F$$

Solution

Given, $$C=\cfrac { { \varepsilon }_{ 0 }A }{ d } =2\mu F$$
and $$C'=\cfrac { K{ \varepsilon }_{ 0 }A }{ d' } =\cfrac { 2.8{ \varepsilon }_{ 0 }A }{ d/2 } =\cfrac { 5.6{ \varepsilon }_{ 0 }A }{ d } $$
$$\Rightarrow C'=5.60\times 2=11.2\mu F$$
Question 10
1 / -0

Two capacitors of capacitance $$C$$ are connected in series. If one of them is filled with a substance of dielectric constant $$K$$, what is the effective capacitance?

A
$$\cfrac { KC }{ \left( 1+k \right) } $$

B
$$C(K+1)$$

C
$$\cfrac { 2KC }{ \left( 1+k \right) } $$

D
None of these

Solution

When a substance of dielectric constant $$K$$ is filled in the capacitor, its capacity
$${ C }_{ m }=\cfrac { K{ \varepsilon }_{ 0 }A }{ d } =K{ C }_{ o }$$
Here, $${ C }_{ o }=C$$
$$\therefore { C }_{ m }==KC$$
Now two capacitors of capacities $$KC$$ and $$C$$ are connected in series, therefore their effective capacitance
$$\cfrac { 1 }{ C' } =\cfrac { 1 }{ KC } +\cfrac { 1 }{ C } =\cfrac { 1+K }{ KC } \Rightarrow C'=\cfrac { KC }{ (1+K) } $$

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Electrostatic Potential and Capacitance Test - 49

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Electrostatic Potential and Capacitance Test - 49

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Question 1

$$\displaystyle\frac{3}{14}\mu F$$

$$\displaystyle\frac{14}{3}\mu F$$

$$21\mu F$$

$$23\mu F$$

Solution

Question 2

The potential difference as well as charge on each capacitor goes up by a factor $$\displaystyle { \varepsilon }_{ r }$$

The potential difference as well as charge on each capacitor goes down by a factor $$\displaystyle { \varepsilon }_{ r }$$

The potential difference across A remains constant and the charge on B remains unchanged

The potential difference across B remains constant while the charge on A remains unchanged

Solution

Question 3

$$2\mu F$$

$$4\mu F$$

$$10\mu F$$

$$12\mu F$$

$$24\mu F$$

Solution

Question 4

$$9\times 10^3$$J

$$9\times 10^9$$J

$$9\times 10^{-3}$$J

$$9\times 10^{-3}$$eV

Solution

Question 5

0.34 m

1.33 m

13.4 m

33.9 m

Solution

Question 6

$$10\ V$$

$$7.5\ V$$

$$5\ V$$

$$2.5\ V$$

Solution

Question 7

Decreases the electric field between the plates

Decreases the capacity of the condenser

Increases the charge stored in the condenser

Increases the capacity of the condenser

Solution

Question 8

$$\displaystyle 11.2\mu F$$

$$\displaystyle 15.6\mu F$$

$$\displaystyle 19.2\mu F$$

$$\displaystyle 22.4\mu F$$

Solution

Question 9

$$11.2\mu F$$

$$15.6\mu F$$

$$19.2\mu F$$

$$22.4\mu F$$

Solution

Question 10

$$\cfrac { KC }{ \left( 1+k \right) } $$

$$C(K+1)$$

$$\cfrac { 2KC }{ \left( 1+k \right) } $$

None of these

Solution

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