Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 36

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Question 1
1 / -0

An air-filled parallel plate capacitor has a capacity $$2 pF$$. The separation between the plates is doubled and the interspace is filled with wax, If the capacity is increased to 6 pF, the dielectric constant of the wax is :

A
$$2$$

B
$$4$$

C
$$3$$

D
$$6$$

Solution

when Air filled , capacitance $$=C=2 pF$$
As C is inversely proportional to the separation so it is double the capacitance will be $$C/2=$$
If dielectric constant is $$k$$, after filling wax the capacitance becomes $$C'=k\dfrac{C}{2}$$
now $$k=\dfrac{C'}{C/2}=\dfrac{6}{2/2}=6 $$
Question 2
1 / -0

Find $$C_{AB}$$ in the given circuit of given figure. Assume each capacitor is C:

A
$$\dfrac{15C}{8}$$

B
$$\dfrac{15C}{7}$$

C
$$\dfrac{3C}{2}$$

D
$$2C$$

Solution

From equivalent circuit as shown in the figure
$$C_{AB}=\frac{3C}{8}+\frac{C}{2}=\frac{15C}{8}$$
Question 3
1 / -0

If each capacitor is C find $$C_{AB}$$ in the given circuit :

A
3C/4

B
4 C

C
C/4

D
C/2

Solution

equivalent circuit of Fig 20.41(a) is
$$\displaystyle C_{AB}=\frac{C\times 3C}{C+ 3 C}=\frac{3C}{4}$$ shown.
Question 4
1 / -0

A capacitor has charge $$50\mu$$ C. When the gap between the plates is filled with glass wool 120 $$\mu $$ C charge flows through the battery. The dielectric constant of glass wool is :

A
$$3.4$$

B
$$1.4$$

C
$$2.4$$

D
none of these

Solution

Before filled wool the charge on capacitor is $$Q$$
After filled wool the charge on capacitor is $$Q'=kQ$$ where $$k=$$ dielectric constant.
So charge flow through the battery is $$Q'-Q=120 \Rightarrow Q'=120+Q=120+50=170$$
thus,$$kQ=170 \Rightarrow k=\dfrac{170}{50}=3.4$$
Question 5
1 / -0

A $$1\ mm$$ thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

A
100

B
100000

C
400000

D
25000

Solution

The electric field in between the plates in the presence of dielectric is
$$E=\displaystyle\frac{V}{kd}=\displaystyle\frac{100}{4\times 10^{-3}}=25000 $$V/m
Question 6
1 / -0

A parallel plate condenser with plate separation d is charged with the help of battery so that $$V_0$$ energy is stored in the system. The battery is now removed. a plate of dielectric constant k and thickness d is placed between the plates of condenser. The new energy of the system will be :

A
$$V_0 k^{-2}$$

B
$$k^2 V_0$$

C
$$V_0 k^{-1}$$

D
$$k V_0$$

Solution

As the battery is removed from the capacitor so the charge (Q) remains constant.
If the capacitance of capacitor before inserting the dielectric is $$C$$, the capacitance of capacitor after inserting the dielectric is $$kC$$
Here, Energy $$V_0=\dfrac{Q^2}{2C}$$
After inserting dielectric the energy becomes $$U=\dfrac{Q^2}{2kC}=V_0k^{-1}$$
Question 7
1 / -0

Two condensers each of capacitance $$2\mu F$$ are connected in parallel and this combination is connected in series with a $$12\mu F$$ capacitor. The resultant capacity of the system will be:

A
$$16\mu f$$

B
$$13\mu f$$

C
$$6\mu f$$

D
$$3\mu f$$

Solution

When two capacitors of capacitance $$2 \mu F$$ are in parallel , the equivalent capacitance is
$$C_{eq}=2+2=4 \mu F$$. Now it is connected to $$12 \mu F$$ in series . So the capacitance of system is $$C_{sys}=\dfrac{4\times 12}{4+12}=3 \mu F$$
Question 8
1 / -0

A parallel plate condenser is connected to a battery of emf 4 volt. If a plate of dielectric constant 8 is inserted into it, the potential difference on the condenser will be :

A
32 V

B
4 V

C
1/2 V

D
2 V

Solution

Total charged is conserved in the condenser. Here the potential difference on the condenser is equal to the emf of battery. When dielectric is inserted between the plates the charge will maintain the constant potential in the capacitor. Thus, the potential difference on the condenser is $$4 V$$
Question 9
1 / -0

A battery of 100 V is connected to series combination of two identical parallel-plate condensers. If dielectric of constant 4 is slipped between the plates of second condenser, then the potential difference on the condensers will respectively become :

A
80 V, 20 V

B
75 V, 25 V

C
50 V, 80 V

D
20 V, 80 V

Solution

Let the capacitance of each capacitors before applied dielectric is $$C$$.
When dielectric is inserted in the second capacitor , the capacitance of first capacitor will remain same i.e $$C_1=C$$ and the capacitance of second capacitor becomes $$C_2=4C$$
As they are in series, equivalent capacitance, $$C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{C. 4C}{C+4C}=\dfrac{4}{5}C$$
Equivalent charge on the circuit is $$Q_{eq}=C_{eq}V=\dfrac{4}{5}C\times 100=80C$$
Now the potential difference on first condenser is $$V_1=\dfrac{Q_{eq}}{C_1}=\dfrac{80C}{C}=80 V$$
Now the potential difference on second condenser is $$V_2=\dfrac{Q_{eq}}{C_2}=\dfrac{80C}{4C}=20 V$$
Question 10
1 / -0

The distance between the plates of a parallel plate air condenser is d. if a copper plate of the same area but thickness $$\dfrac d2$$ is placed between the plates then the new capacitance will become :

A
doubled

B
half

C
one fourth

D
remain unchanged

Solution

Capacitance of parallel plate capacitor is $$C=\dfrac{A\epsilon}{d}$$
If distance between plate is $$d/2$$
so new capacitance is $$C'=\dfrac{A\epsilon}{d/2}=2\dfrac{A\epsilon}{d}=2C$$