Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

In the given network of capacitors as shown in figure, given that $${ C }_{ 1 }=C_{ 2 }={ C }_{ 3 }=400pF$$ and $$C_{ 4 }={ C }_{ 5 }={ C }_{ 6 }=200pF$$. The effective capacitance of the circuit between $$X$$ and $$Y$$ is :

A
$$810pF$$

B
$$205pF$$

C
$$600pF$$

D
$$410pF$$

Solution

Let equivalent capacitance between X and Y be
$${ C }_{ eq }\quad '\parallel '\longrightarrow $$parallel combination $$'+'\longrightarrow $$
$$\therefore { C }_{ eq }=\left\{ \left[ \left( { C }_{ 3 }\parallel { C }_{ 4 }+{ C }_{ 2 } \right) \parallel { C }_{ 5 } \right] +{ C }_{ 1 } \right\} \parallel { C }_{ 6 }$$
$$\therefore { C }_{ eq }=\left\{ \cfrac { \left[ \cfrac { \left( { C }_{ 3 }+{ C }_{ 4 } \right) \times { C }_{ 2 } }{ { C }_{ 3 }+{ C }_{ 4 }+{ C }_{ 2 } } +{ C }_{ 5 } \right] { C }_{ 1 } }{ \cfrac { \left( { C }_{ 3 }+{ C }_{ 4 } \right) \times { C }_{ 2 } }{ { C }_{ 3 }+{ C }_{ 4 }+{ C }_{ 2 } } +{ C }_{ 5 }+{ C }_{ 1 } } \right\} +{ C }_{ 6 }$$
By substituting, $${ C }_{ 1 }={ C }_{ 2 }={ C }_{ 3 }=400pF\& { C }_{ 4 }={ C }_{ 5 }={ C }_{ 6 }=200pF$$
We get, $${ C }_{ eq }=410pF$$ Thus option D
Question 2
1 / -0

An uncharged parallel plate capacitor having a dielectric of dielectric constant $$K$$ is connected to a similar air cored parallel plate capacitor charged to a potential $${V}_{0}$$. The two share the charge, and the common potential becomes $$V$$. The dielectric constant $$K$$ is :

A
$$\cfrac { { V }_{ 0 } }{ V } -1$$

B
$$\cfrac { { V }_{ 0 } }{ V } +1$$

C
$$\cfrac { V }{ { V }_{ 0 } } -1$$

D
$$\cfrac { V }{ { V }_{ 0 } } +1$$

Solution
Question 3
1 / -0

Two parallel plate capacitors of capacitances $$C$$ and $$2C$$ are connected in parallel and charged to a potential difference $$V$$. The battery is then disconnected, and the region between the plates of $$C$$ is filled completely with a material of dielectric constant $$K$$. The common potential difference across the combination becomes :

A
$$\cfrac {2V}{K+2}$$

B
$$\cfrac {V}{K+1}$$

C
$$\cfrac {3V}{K+3}$$

D
$$\cfrac {3V}{K+2}$$

Solution

Initially both capacitors are at potential $$V$$.
Charge on both plates are $$CV$$ and $$2CV$$.
When we put dielectric between the plates of $$C$$,
$$C'=KC$$
Common potential, $${ V }_{ C }=\cfrac { { Q }_{ 1 }+{ Q }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } =\cfrac { CV+2CV }{ KC+2C } =\cfrac { 3V }{ K+2 } $$
Question 4
1 / -0

The equivalent capacitance of the circuit across the terminals $$A$$ and $$B$$ is equal to :

A
$$0.5\mu F$$

B
$$2\mu F$$

C
$$1\mu F$$

D
none of these

Solution

As shown in the figure, both capacitors $$ C=2\mu F$$ are in parallel so $$C_{1}=2+2=4\mu F$$
Now, capacitors $$C=2/3 \mu F $$ are also in parallel combination so $$C_{2}=2/3+2/3=4/3 \mu F$$
$$C_1$$ and $$C_2$$ are now in series combination
so $$C_{eq}=\dfrac{C_1C_2}{C_1+C_2 }=1\mu F$$
Question 5
1 / -0

The plates of a parallel plate capacitor are charged upto $$100V$$. now, after removing the battery, a $$2mm$$ thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by $$1.6mm$$. The dielectric constant of the plate is :

A
$$5$$

B
$$1.25$$

C
$$4$$

D
$$2.5$$

Solution

As battery is disconnected, so charge will remain the same.
It is given that final potential is the same.
So final capacitance should be $$C_1=C_2$$
$$\cfrac { { \varepsilon }_{ 0 }A }{ d } =\cfrac { { \varepsilon }_{ 0 }A }{ (1.6+d)-t(1-1/K) } \Rightarrow K = 5 $$
Question 6
1 / -0

In the figure, identical capacitors are connected in the given three configurations. The ratio of the total capacitances in (i), (ii) and (iii) respectively, is :

A
$$3:5:5$$

B
$$3:3:5$$

C
$$5:4:4$$

D
$$5:5:3$$

Solution

The equivalent circuits of each network are shown in figure below.
For (a), Total capacitance is $$C_a=C+\dfrac{C(C+C)}{C+(C+C)} =C+2C/3=(5/3)C$$
For (b), Total capacitance is $$C_b=C+\dfrac{C(C+C)}{C+(C+C)} =C+2C/3=(5/3)C$$
For (c), As this is a wheatstone bridge circuit, total capacitance is $$C_c=C$$
Thus, $$C_a:C_b:C_c=(5/3)C:(5/3)C:C=5:5:3$$
Question 7
1 / -0

Consider the figure given. Each capacitor has capacitance $$C$$. The equivalent capacitance between $$1$$ and $$3$$ is :

A
0

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

C
$$\cfrac{5C}{2}$$

D
$$\cfrac{5C}{4}$$

Solution

Here 2,4 and 5 are equipotential points and the equivalent circuit is shown in below.
The equivalent capacitance between 1 and 3 is $$C_{13}=\dfrac{(C+C+C)(C+C+C)}{(C+C+C)+(C+C+C)}+C=\dfrac{(3C) (3C)}{3C+3C}+C=\dfrac{3C}{2}+C=\dfrac{5C}{2}$$
Question 8
1 / -0

An electric charge $$10^{-3}\mu C$$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at $$(\sqrt 2, \sqrt 2)$$ and (2, 0) respectively. The potential difference between the points A and B will be :

A
4.5 volt

B
9 volt

C
zero

D
2 volt

Solution

The distance of point A from origin is $$r_A=\sqrt{(\sqrt 2-0)^2+(\sqrt 2-0)^2}=\sqrt{4}=2$$
The distance of point B from origin is $$r_B=\sqrt{(2-0)^2+(0-0)^2}=\sqrt{4}=2$$
Potential at A due to charge $$Q=10^{-3} \mu C$$ is $$\displaystyle V_A=\frac{Q}{4\pi\epsilon_0 r_A}=\frac{Q}{8\pi\epsilon_0 }$$ as $$r_A=2$$
and potential at B is $$\displaystyle V_B=\frac{Q}{4\pi\epsilon_0 r_B}=\frac{Q}{8\pi\epsilon_0 }$$ as $$r_B=2$$
Potential difference between A and B $$\displaystyle=V_A-V_B=\frac{Q}{8\pi\epsilon_0 }-\frac{Q}{8\pi\epsilon_0 }=0$$ volt.
Question 9
1 / -0

Consider the figure given. Each capacitor has capacitance $$C$$. The capacitance between $$1$$ and $$3$$ is :

A
$$\cfrac{3C}{4}$$

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

C
$$\cfrac{5C}{2}$$

D
$$\cfrac{5C}{4}$$

Solution

Since 1, 2 and 3 are equipotential, we can consider them as single point, say 6. Now there is effectively 3 capacitances lying parallel between 4 & 6 (3C) & another 3 capacitances lying parallel between 6 & 5 (3C) & the two parallel combination being in series (3C/2).
Thus, the net capacitance is: $${C}_{t}=\dfrac{1}{\dfrac{1}{C+C+C}+\dfrac{1}{C+C+C}}=\dfrac{3C}{2}$$
Question 10
1 / -0

An air capacitor of capacity $$C=10\mu F$$ is connected to a constant voltage battery of $$12$$ volt. Now the space between the plates is filled with a liquid of dielectric constant $$5$$. The (additional) charge that flows now from battery to the capacitor is :

A
$$120 \mu C$$

B
$$600 \mu C$$

C
$$480 \mu C$$

D
$$24 \mu C$$

Solution

Initial charge $$q_i=CV=10(12)=120 \mu C$$.
After filled with dielectric liquid the charge $$q'=C'V=KCV=5(10)12=600 \mu C$$
The (additional) charge that flows now from battery to the capacitor $$=q_f-q_i=600-120=480 \mu C$$