Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 44

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Question 1
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A dielectric is inserted into a capacitor while the charge on it is kept constant. What happens to the potential difference and the stored energy?

A
The potential difference decreases and the stored energy increases

B
Both the potential difference and the stored energy increase

C
The potential difference increases and the stored energy decreases

D
Both the potential difference and the stored energy decrease

E
Both the potential difference and the stored energy remain the same

Solution

When a dielectric is placed in an electric field then an internal electric field is produced in it due to polarisation of dielectric . This internal electric field inside the dielectric is in opposite direction of the field between plates of capacitor , as a result of this effective electric field between plates decreases , hence the potential difference between plates because ,
$$E=\Delta V/\Delta r$$ ,
where $$\Delta r $$ is the distance between plates , which is constant .
Now potential energy of a capacitor is given by ,
$$U=(1/2)q\Delta V$$ ,
charge is given constant , therefore due to decrease in potential difference $$\Delta V$$ , potential energy $$U$$ also decreases .
Question 2
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Below, three different arrangement of identical capacitors are shown. What is the correct order of total capacitance values for the arrangements, greatest first?

A
$$1, 2, 3$$

B
$$3, 2, 1$$

C
$$3, 1, 2$$

D
$$2, 3, 1$$

E
$$2, 1, 3$$

Solution

Let the capacitance of each capacitor be $$C$$.
Case 1 : Capacitance $$C_1 = C$$
Case 2 : Equivalent capacitance of series connection $$C_2 = \dfrac{C\times C}{C+C} = \dfrac{C}{2}$$

Case 3 : Equivalent capacitance of parallel connection $$C_3= C+C = 2C$$
$$\implies$$ $$C_3>C_1>C_2$$
Question 3
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A negative charge is located in the uniform electric field between two plates (not shown).
Where in the diagram will the electron have the greatest change in potential energy before hitting the one of the plates?

A
A

B
B

C
C

D
D

Solution

A charged species tends to move from place of higher potential energy to that of lower potential energy.
The direction of electric field is from left to right. This means that there is a positively charged plate on the left side, closest to A and farthest from B.
An electron being negatively charged plate always moves towards left here. Hence the maximum potential energy resides at B.
Question 4
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A $$5F$$ capacitor holds $$100C$$ of charge when it is connected to a power supply. The $$5F$$ capacitor is removed and replaced with two $$40F$$ capacitors in series with each other, the power supply remains untouched.
What is the new charge stored in one of the two $$40F$$ capacitors?

A
$$100C$$

B
$$400C$$

C
$$800C$$

D
$$1200C$$

Solution

Given : $$C_i =5F$$ $$Q_i = 100$$ C
Thus voltage of power supply $$V = \dfrac{Q_1}{C_1} = \dfrac{100}{5} =20$$ volts

Now the $$5F$$ is replaced by two $$C =40F$$ capacitors connected in series.
Equivalent capacitance of two capacitors $$C_{eq} = \dfrac{C \times C}{C+C} = \dfrac{C}{2} = \dfrac{40}{2} =20F$$
$$\therefore$$ New charge stored $$Q' = C{_eq}V = 20 \times 20 = 400$$ C
Question 5
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Each capacitor in the circuit shown is a $$1F$$ capacitor.
What potential across $$A$$ and $$B$$ would result in $$2$$ Coulombs of charge on each plate of each capacitor?

A
$$2V$$

B
$$4V$$

C
$$6V$$

D
$$8V$$

Solution

Equivalent capacitance of each pair of capacitance in series=$$\dfrac{1\times 1}{1+1}F=0.5F$$
Total charge stored in such a combination=$$2C+2C=4C$$
Potential difference across such combination=$$V$$
Hence $$V=\dfrac{Q}{C}=\dfrac{4C}{0.5F}=8volt$$
Question 6
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Each capacitor in the circuit shown is a $$1F$$ capacitor. What would be the equivalent capacitance between $$A$$ and $$B$$?

A
$$0.5F$$

B
$$F$$

C
$$2F$$

D
$$4F$$

Solution

Equivalent capacitance of each pair of capacitance in series=$$\dfrac{1\times 1}{1+1}F=0.5F$$
The two series combination are connected in parallel. Hence the net capacitance becomes $$0.5F+0.5F=1F$$
Correct answer is option B.
Question 7
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A student is working on a circuitry project and wants to create an equivalent capacitance of $$700F$$. However, the student's only supply is a bag of $$470F$$ capacitors.
How can the student arrange any number of the $$470F$$ capacitors so that their equivalent capacitance is as close to $$700F$$ as possible?

A
Two capacitors in parallel.

B
Two capacitors in series.

C
Three capacitors in parallel.

D
Two capacitors in series connected to one other in parallel.

E
Two capacitors in parallel connected to one other in series.

Solution

Given : $$C = 470$$ F
Equivalent capacitance of capacitors connected in series $$C_s = \dfrac{C \times C}{C+C} = \dfrac{C}{2} = 235$$ F

$$\therefore$$ Equivalent capacitance of the whole circuit $$C_{eq} = C+ C_s = 470+ 235 = 705$$ F $$\approx 700$$ F
Question 8
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An air-filled capacitor has a capacitance of $$C^o$$. Then a dielectric with a dielectric constant of 3.0 is inserted into it and the plates are pushed toward each other so that the distance between the plates is reduced to $$\frac{1}{2}$$ the original value.
What is the new capacitance?

A
$$\frac{1}{6}C^o$$

B
$$\frac{2}{3}C^o$$

C
$$\frac{3}{2}C^o$$

D
$$6C^o$$

Solution

Capacitance of air-filled parallel plate capacitor $$C^o =\dfrac{A\epsilon_o}{d}$$
Dielectric constant of the inserted dielectric $$K = 3.0$$
New distance between the plates $$d' =d/2$$

$$\therefore$$ New capacitance $$C' = \dfrac{KA\epsilon_o}{d'} = \dfrac{(3.0)A\epsilon_o }{d/2} = 6C^o$$
Question 9
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A capacitor contains two square plates with side lengths $$5.0$$ cm. The plates are separated by $$2.0$$ mm. Dry air fills the space between the plates. Dry air has a dielectric constant of $$1.00$$ and experiences dielectric breakdown when the electric field exceeds $$3.010^4 V/cm$$.
What is the magnitude of charge that can be stored on each plate before the capacitor exceeds its breakdown limit and sends a spark between the plates?

A
$$6.6\times 10^{-8}C$$

B
$$6.6\times 10^{-5}C$$

C
$$3.3\times 10^{-7}C$$

D
$$3.3\times 10^{-8}C$$

E
$$8.1\times 10^{-2}C$$

Solution

We have $$C=Q/V$$
or $$Q=CV$$ ,
for parallel plate capacitor , $$C=\varepsilon_{0} KA/d$$ ,
where $$A=$$ area of a plate ,
$$d=$$ separation between plates ,
and $$E=V/d$$ or $$V=Ed$$
therefore $$Q=\varepsilon_{0}KA/d\times Ed$$ ,
or $$Q=\varepsilon_{0}KA\times E$$
now charge on a plate $$Q$$ will be maximum when electric field $$E$$ is maximum ,
maximum value of $$E=3.0\times 10^{4} V/cm=3.0\times10^{6}V/m$$ ,
area of a plate $$A=5.0^{2}=25cm^{2}=25\times10^{-4}m^{2}$$ ,
therefore maximum charge ,
$$Q=8.854\times10^{-12}\times1\times25\times10^{-4}\times3.0\times10^{6}$$
or $$Q=6.6\times10^{-8}C$$
Question 10
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Three identical capacitors, each of capacitance $$C$$, are wired as shown above
When both switches are closed, the equivalent capacitance is:

A
$$\cfrac{C}{3}$$

B
$$\cfrac{2}{3}C$$

C
$$C$$

D
$$\cfrac{3}{2}C$$

E
$$3C$$

Solution

As capacitors $$C_{2}$$ and $$C_{3}$$ are in parallel combination, therefore, equivalent capacitance of these two,
$$C'=C_{2}+C_{3}=C+C=2C$$

Now $$C_{1}$$ and $$C'$$ are in series, therefore
$$C=\frac{C_{1}\times C'}{C_{1}+C'}$$
$$C=\frac{C\times 2C}{C+2C}=2C^{2}/3C=2C/3$$