Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 73

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Question 1
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A parallel plate capacitor has a capacity $$80\times 10^{-6}\ F$$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $$20$$. The capacitor is now connected to a battery of $$30\ V$$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :

A
$$45.6\times 10^{-3}\ C$$

B
$$25.3\times 10^{-3}\ C$$

C
$$120\times 10^{-3}\ C$$

D
$$12\times 10^{-3}\ C$$

Solution

Charge passing through the wire, $$\Delta q=\Delta CV$$
$$=(C'-C)V\\=(k-1)C V\\=(20-1)(80\times 10^{-6})(30)\\=45.6\times 10^{-3}$$
So, charge passing through the wire $$=45.6 \times 10^{-3}\ C$$
Question 2
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Consider a parallel plate capacitator of capacity 10 $$\displaystyle \mu F$$ filled with air. When the gap between the plates is filled partly with a dielectric of dielectric constant $$4$$, as shown in figure, the new capacity of the capacitator is ($$A$$ is the area of plates):

A
$$20 \displaystyle \mu F$$

B
$$40\displaystyle \mu F$$

C
$$2.5\displaystyle \mu F$$

D
$$25\displaystyle \mu F$$

Solution

Such a configuration can be thought of as a parallel combination of two capacitors- one with area A/2 and dielectric and another with same area and without dielectric.

$${ C }_{ 1 }=\dfrac { { \epsilon }_{ 0 }A }{ 2d } $$; (for air in between)

In second case $$k = 4$$

$${ C }_{ 2 }=\dfrac { 4{ \epsilon }_{ 0 }A }{ 2d } =\dfrac { 2{ \epsilon }_{ 0 }A }{ d } $$

Hence, effective capacitance is $$C={ C }_{ 1 }+{ C }_{ 2 }=2.5\dfrac { { \epsilon }_{ 0 }A }{ d } =2.5{ C }_{ 0 }=25\ \mu F$$
Question 3
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Two identical condensers M and N are connected in series with a battery. The space between the plates of M is completely filled with a dielectric medium of dielectric constant $$8$$ and a copper plate of thickness $$\displaystyle \frac { d }{ 2 } $$ is introduced between the plates of N ($$d$$ is the distance between the plates). Then potential differences across M and N are, respectively, in the ratio:

A
$$1:4$$

B
$$4:1$$

C
$$3:8$$

D
$$1:6$$

Solution

Let the capacitance of the condenser $$M$$ be $$C$$ and $$N$$ be $$C'$$.

The new capacitance of the condenser $$M$$ will be, $$C_M=\varepsilon C_M=8C$$

When a copper plate of thickness $$d/2$$ is inserted in between the capacitor N, effective distance between plates reduces to $$d-d/2=d/2$$.
Since, $$C\propto \dfrac{1}{d}, \quad C'=2C$$

Since, the two condensers are connected in series, charge one each capacitor is same.
$$\Rightarrow V_M:V_N=\dfrac{1}{C_M}:\dfrac{1}{C_N}=\dfrac{1}{8C}:\dfrac{1}{2C}$$
$$\Rightarrow V_M:V_N=1:4$$
Question 4
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A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure. The new capacitance will be :

A
$$\dfrac{(K + 3) C}{4}$$

B
$$\dfrac{(K + 2) C}{4}$$

C
$$\dfrac{(K + 1) C}{4}$$

D
$$\dfrac{KC}{4}$$

Solution

Capacitance $$=\dfrac{\epsilon A}{d}$$

Initially for air: $$C=\dfrac{\epsilon_o A}{d}$$

Finally for air: $$C=\dfrac{\epsilon_o \dfrac{3A}{4}}{d}=\dfrac{3C}{4}$$

For dielectric: $$C_k=\dfrac{K\epsilon_o \dfrac{A}{4}}{d}=\dfrac{KA}{4}$$

We can consider them as two capacitors in series which gives,
$$C_{net}=\dfrac{3C}{4}+\dfrac{KC}{4}=\dfrac{(K+3)C}{4}$$
Question 5
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Two metal plates each of area $$A$$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $$d$$. A metal plate of thickness $$\dfrac{d}{2}$$ and of same area $$A$$ is inserted between the plates to form two capacitors of capacitances $$C_1$$ and $$C_2$$ as shown in the figure. The effective capacitance of the two capacitors is C' and the capacitance of the capacitor initially is C, then $$\frac{C'}{C}$$ is :

A
$$4$$

B
$$2$$

C
$$6$$

D
$$1$$

Solution

Let the gap for capacitor 1 be x and that for capacitor 2 be y. Then,
$${ C }_{ 1 }=\dfrac { { \epsilon }_{ 0 }A }{ x } $$ and $${ C }_{ 2 }=\dfrac { { \epsilon }_{ 0 }A }{ y } $$
$$x+y=d/2$$

Since the two are in series,
$$C'=\dfrac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } ={ \epsilon }_{ 0 }A\dfrac { \dfrac { 1 }{ xy } }{ \dfrac { 1 }{ x } +\dfrac { 1 }{ y } }$$

$$ =\dfrac { { \epsilon }_{ 0 }A }{ x+y } =\dfrac { 2{ \epsilon }_{ 0 }A }{ d } =2C$$

Hence , $$\dfrac{C'}{C}=2$$
Question 6
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The potential difference between two points A and B is $$10V$$. Point A is at higher potential. If a negative charge, q = 2 C, is moved from point A to point B, then the potential energy of this charge will:

A
decrease by 20 J

B
decrease by 5 J

C
increase by 5 J

D
increase by 20 J

E
increase by 100 J

Solution

The change in potential energy, $$\Delta U=q\Delta V$$
As the charge $$q=-2C$$ is moved from A to B so potential will decrease 10 V i.e $$\Delta V=-10 V$$
Thus, $$\Delta U=(-2)(-10) =+20 J$$
Question 7
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Five capacitors, each of capacitance value $$C$$ are connected as shown in the figure. The ratio of capacitance between $$P$$ & $$R$$, and the capacitance between $$P$$ & $$Q$$, is :

A
$$3 : 1$$

B
$$5 : 2$$

C
$$2 : 3$$

D
$$1 : 1$$

Solution

Hint: reduce the circuit into simple form and find equivalent capacitance

Step 1: find equivalent capacitance between P&R

In series combination,

$$\dfrac{1}{{{C}_{1}}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C} $$

$$ {{C}_{1}}=\dfrac{C}{3} $$

$$\dfrac{1}{{{C}_{2}}}=\dfrac{1}{C}+\dfrac{1}{C} $$

$${{C}_{2}}=\dfrac{C}{2}$$

In parallel combination,

C(parallel) = C(1) + C(2)

$$ {{C}_{pr}}=\dfrac{C}{3}+\dfrac{C}{2} $$

$$ {{C}_{pr}}=\dfrac{5C}{6} $$

Step 2:
find equivalent capacitance between P&Q

$$ \dfrac{1}{{{C}_{1}}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C} $$

$${{C}_{1}}=\dfrac{C}{4} $$

$${{C}_{pq}}=\dfrac{C}{4}+C $$

$$ {{C}_{pq}}=\dfrac{5C}{4} $$
and,

$$ {{C}_{pr}}=\dfrac{5C}{6} $$

$$ {{C}_{pq}}=\dfrac{5C}{4} $$

$$\dfrac{{{C}_{pr}}}{{{C}_{pq}}}=\dfrac{5C}{6}\times \dfrac{4}{5C} $$

$$ \dfrac{{{C}_{pr}}}{{{C}_{pq}}}=\dfrac{2}{3} $$

Step 3: find the ratio

$${{C}_{pr}}=\dfrac{5C}{6} $$

$$ {{C}_{pq}}=\dfrac{5C}{4} $$

$$ \dfrac{{{C}_{pr}}}{{{C}_{pq}}}=\dfrac{5C}{6}\times \dfrac{4}{5C} $$

$$ \dfrac{{{C}_{pr}}}{{{C}_{pq}}}=\dfrac{2}{3} $$
Final step: The ratio of capacitance between P&R
, and the capacitance between P&Q, is 2:3.
Question 8
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In the given figure, the capacitors $${ C }_{ 1 },{ C }_{ 3 },{ C }_{ 4 },{ C }_{ 5 }$$ have a capacitance $$4 \mu F$$ each. If the capacitor $${ C }_{ 2 }$$ has a capacitance $$10 \mu F$$, then effective capacitance between $$A$$ and $$B$$ will be

A
$$2 \mu F$$

B
$$6 \mu F$$

C
$$4 \mu F$$

D
$$8 \mu F$$

Solution

When a battery is applied across $$A$$ and $$B$$, then the points $$b$$ and $$c$$ will be at the same potential $$\left( \because { C }_{ 1 }={ C }_{ 4 }={ C }_{ 3 }={ C }_{ 5 }=4\mu F \right)$$
Therefore, no charge flows through $$ { C }_{ 2 }$$.
As, $${ C }_{ 1 }$$ and $${ C }_{ 5 }$$ are in series.
$$\therefore$$ Their equivalent capacitance,
$$ { C }^{ \prime }=\dfrac { { C }_{ 1 }\times { C }_{ 5 } }{ { C }_{ 1 }+{ C }_{ 5 } } =\dfrac { 4\times 4 }{ 4+4 } =2\mu F$$
Similarly, $${ C }_{ 4 }$$ and $${ C }_{ 3 }$$ are in series. Therefore, their equivalent capacitance
$${ C }^{ \prime \prime }=\dfrac { { C }_{ 3 }\times { C }_{ 4 } }{ { C }_{ 3 }+{ C }_{ 4 } } =\dfrac { 4\times 4 }{ 4+4 } =2\mu F$$
Now, $${ C }^{ \prime }$$ and $${ C }^{ \prime \prime }$$ are in parallel. Therefore, effective capacitance between $$A$$ and $$B$$
$$={ C }^{ \prime }+{ C }^{ \prime \prime }=2+2=4\mu F$$
Question 9
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A particle of mass $$10^{-3}kg$$ and charge $$5\mu C$$ is thrown at a speed $$20\ m\ s^{-1}$$ against a uniform electric field of strength $$2\times 10^{5}N\ C^{-1}$$. How much distance will it travel before coming to rest momentarily?

A
$$0.1\ m$$

B
$$0.3\ m$$

C
$$0.05\ m$$

D
$$0.2\ m$$

Solution

Force on particle$$=F=qE$$ in opposite direction of motion

And, $$F=ma=qE$$

$$\implies a=\dfrac{qE}{m}=\dfrac{5\times 10^{-6}\times 2\times 10^5}{10^{-3}}$$

$$\implies a=10^3m/s^2$$ and this acceleration is negative since particle is thrown against force.
And final velocity is $$v=0$$

Using $$v^2-u^2=2as$$

$$\implies 0-20^2=-2\times 1000\times s$$

$$\implies s=0.2m$$

Answer-(D)
Question 10
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The V versus x plot for six identical metal plates of cross-sectional area A is as shown.The equivalent capacitance between 2 and 5 is (Adjacent plates are placed at a separation d):

A
$$\dfrac{2 \varepsilon _0 A}{d}$$

B
$$\dfrac{\varepsilon _0 A}{d}$$

C
$$\dfrac{3 \varepsilon _0 A}{d}$$

D
$$\dfrac{\varepsilon _0 A}{2d}$$

Solution