Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 22

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Question 1
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Effective capacitance of parallel combination of two capacitors $$C_1$$ and $$C_2$$ is $$10\mu F$$. when these capacitors are individually connected to a voltage source of $$1V$$, the energy stored in the capacitor $$C_2$$ is 4 times that of $$C_1$$. If these capacitors are connected in series, their effective capacitance will be:

A
$$8.4\mu F$$

B
$$3.2\mu F$$

C
$$1.6\mu F$$

D
$$4.2\mu F$$

Solution

Given that $$C_1 + C_2 = 10\mu F$$ (Parallel combination) ...(1)

They are now connected to the same voltage source of $$1V$$.

The energy stored in the two capacitors are given by

$$E_1 = \dfrac{1}{2}C_1V^2$$ $$E_2 = \dfrac{1}{2}C_2V^2$$

Given that $$E_2 = 4E_1$$

$$\therefore \dfrac{1}{2} C_2V^2 = 4 \times C_1V^2$$

$$\therefore C_2 = 4C_1$$ ....(ii)

From (i) and (ii),

$$C_1 + C_2 = C_1 + 4C_1 = 10\mu F$$.

$$5C_1 = 10\mu F$$, $$C_1 = 2\mu F$$

$$C_2 = 10\mu F-C_1 = 10 \mu F - 2\mu F = 8\mu F$$.

When $$C_1$$ and $$C_2$$ are connected in series, the equivalent capacitance of the combination is given by

$$C_{series} = \dfrac{C_1C_2}{C_1+C_2} = \dfrac{2\times 8}{2+8}\mu F = 1.6\mu F$$
Question 2
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The equivalent capacitance between $$A$$ and $$B$$ in the circuit given below is:

A
$$4.9\ \mu F$$

B
$$3.6\ \mu F$$

C
$$5.4\ \mu F$$

D
$$2.4\ \mu F$$

Solution

$$C_1 = 5 + 5 + 2 = 12 \mu F$$
$$C_2 = 4 + 2 = 6 \mu F$$
$$1/ C_{Eq} = 1/6 + 1/12 + 1/6 = 5/12$$
$$C_{Eq} = 2.4 \mu F$$
Question 3
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Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $$\mu F$$ is :

A
$$\dfrac{32}{23}\mu F$$

B
$$\dfrac{31}{23}\mu F$$

C
$$\dfrac{33}{23}\mu F$$

D
$$\dfrac{34}{23}\mu F$$
Question 4
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Three capacitances, each of $$3\mu F$$, are provided. These cannot be combined to provide the resultant capacitance of :

A
$$1\mu F$$

B
$$2\mu F$$

C
$$4.5\mu F$$

D
$$6\mu F$$

Solution

We are provided with 3 capacitors each of capacitance 3 $$ \mu F $$
We prove that the rest of the three combinations can be obtained with a combination of these.
$$ 1 \mu F $$ = Combining all three of them in series
$$ 2 \mu F $$ = Combining 2 in parallel and 1 in series with them
$$ 4.5 \mu F $$ = Combining 2 in series and 1 in parallel combination
Hence, only the 6 $$ \mu F $$ combination is not possible
Question 5
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A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

A
decreases

B
remains unchanged

C
becomes infinite

D
increases.

Solution

The capacitance without aluminium foil is $$C=\dfrac{A\epsilon_0}{d}$$

The capacitance with aluminium foil is $$C'=\dfrac{A\epsilon_0}{d-t}$$

If thickness $$(t)$$ of foil is negligible so $$d-t \sim d.$$ Hence, $$C=C'$$
Question 6
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Three capacitors each of $$4\ \mu F$$ are to be connected in such a way that the effective capacitance is $$6\ \mu F$$. This can be done by connecting them:

A
two in parallel and one in series

B
all in parallel

C
all in series

D
two in series and one in parallel

Solution

For capacitor is series, equivalent capacitance is given by:
$$C_{S}=\dfrac{C_1C_2}{C_1+C_2}$$

For capacitor is parallel, equivalent capacitance is given by:
$$C_{P}=C_1+C_2$$

In the given configuration, equivalent capacitance is given by
$$C_{eq}=C_s+C$$, where $$C= 4\ \mu F$$
and $$C_s=\dfrac{C \times C}{C+C}=\dfrac{C}{2}$$

$$\therefore C_{eq}=C+\dfrac{C}{2}=6\ \mu F$$
Question 7
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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
$$9$$ volt

B
zero

C
$$2$$ volt

D
$$4.5$$ volt

Solution

$$ V_A = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_A} = \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{2+2}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2}$$

$$ V_B= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{r_B}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{\sqrt{4+0}}= \dfrac{1}{4 \pi \epsilon } \dfrac{ q}{2} $$

$$ V_A - V_B =0 $$
Question 8
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A parallel plat capacitor is made of two circular plates separated by a distance of $$5 mm$$ and with a dielectric of dielectric constant $$2.2$$ between them. When the electric field in the dielectric is $$3 \times 10^4$$V/m, the charge density of the positive plate will be close to:

A
$$3 \times 10^4 C/m^2$$

B
$$6 \times 10^4 C/m^2$$

C
$$6 \times 10^{-7} C/m^2$$

D
$$3 \times 10^{-7} C/m^2$$

Solution

Answer is C.
By formula of electric field between the plates of a capacitor $$ \displaystyle E = \dfrac{\sigma}{K\varepsilon_0}$$
$$\sigma = EK\varepsilon_0 = 3 \times 10^4 \times 2.2 \times 8.85 \times 10^{-12}$$
$$= 6.6 \times 8. 85 \times 10^{-8}$$
$$= 5.841 \times 10^{-7}$$
$$\cong 6 \times 10^{-7} C/m^2$$
Question 9
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A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $$K = 2.$$The level of liquid is $$\dfrac{d}{3}$$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time $$t$$ is :

A
$$\displaystyle \dfrac{6\epsilon_{0}\mathrm{R}}{5\mathrm{d}+3\mathrm{V}\mathrm{t}}$$

B
$$\displaystyle \dfrac{(15\mathrm{d}+9\mathrm{V}\mathrm{t})\epsilon_{0}\mathrm{R}}{2\mathrm{d}^{2}-3\mathrm{d}\mathrm{V}\mathrm{t}-9\mathrm{V}^{2}\mathrm{t}^{2}}$$

C
$$\displaystyle \dfrac{6\epsilon_{0}\mathrm{R}}{5\mathrm{d}-3\mathrm{V}\mathrm{t}}$$

D
$$\displaystyle \dfrac{(15\mathrm{d}-9\mathrm{V}\mathrm{t})\epsilon_{0}\mathrm{R}}{2\mathrm{d}^{2}+3\mathrm{d}\mathrm{V}\mathrm{t}-9\mathrm{V}^{2}\mathrm{t}^{2}}$$

Solution

At time t the level of liquid is,
$$ \dfrac { d }{ 3 }$$$$ -Vt\\ { C }_{ eq }=\dfrac { \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } \cdot \dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } }{ \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } +\dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } } =\dfrac { { \epsilon }_{ 0 }KA }{ K\left( \dfrac { 2d }{ 3 } +Vt \right) +\dfrac { d }{ 3 } -Vt } \\ =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 4d }{ 3 } +2Vt \right) +\dfrac { d }{ 3 } -Vt } =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 5d }{ 3 } +Vt \right) } =\dfrac { 6{ \epsilon }_{ 0 } }{ 5d+3Vt } \\ \tau =R{ C }_{ eq }=\dfrac { 6{ \epsilon }_{ 0 }R }{ 5d+3Vt } $$
Question 10
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A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

A
The change in energy stored is $$\frac {1}{2}CV^2\left (\frac {1}{K}-1\right )$$

B
The charge on the capacitor is not conserved

C
The potential difference between the plates decreases K times

D
The energy stored in the capacitor decreases K times

Solution

$$C_f = KC_i$$
$$Q_i =C_iV$$
$$U_i = \frac{1}{2}C_iV^2$$
$$U_f = \frac{1}{2}\frac{Q_f ^2}{C_f}$$
After the capacitor is disconnected from the battery, charge on the capacitor cannot change.
$$\therefore Q_f =Q_i$$
$$ \therefore U_f = \frac{1}{2} \frac{Q_f ^2}{C_f}= \frac{1}{2}\frac{C_i ^2V^2}{KC_i}$$
$$ \Delta U = U_f -U_i = \frac{1}{2}\frac{C_i ^2V^2}{KC_i} - \frac{1}{2}C_i V^2= \frac{1}{2}C_i V^2 ( \frac{1}{K} -1)=\frac{1}{2}C V^2 ( \frac{1}{K} -1) $$ since $$C_i= C$$ as given

Option(A) is correct.

Option (B) :
Charge on the capacitor remains same since battery is disconnected.

Option (C):
C increases K times.
Since $$Q = CV$$ and Q remains same, V decreases K times.

Option D:
The energy stored in the capacitor $$U_f$$ decreases K times.