Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 59

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Four identical capacitors are connected such that ,three capacitors are in parallel the fourth is connected in series. The effective capacity is $$3.75$$ $$\mu F$$ then each capacitor is of value

A
$$3\ \mu F$$

B
$$4\ \mu F$$

C
$$5\ \mu F$$

D
$$6\ \mu F$$

Solution

Given that

Effective capacitance $$C=3.75\,\mu F$$

Given that,

Three identical capacitor are connected in parallel

$$ C'=C+C+C $$

$$ C'=3C $$

And fourth is connected in series

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

$$ \dfrac{1}{3.75}=\dfrac{1}{3C}+\dfrac{1}{C} $$

$$ C=5\,\mu F $$

Hence, the value of each capacitor is $$5\,\mu F$$
Question 2
1 / -0

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

A
$$13 \mu F$$

B
$$\dfrac { 36 } { 13 } \mu F$$

C
$$3 \mu F$$

D
$$\dfrac { 3 } { 4 } \mu F$$

Solution

$$\therefore {C_{eq}} = \frac{{9 \times 4}}{{9 + 3}} = \frac{{36}}{{13}}\mu F$$
$$\therefore$$ Option $$B$$ is correct.
Question 3
1 / -0

Two identical parallel plate capacitors of same dimensions are connected to a D.C. source in series When one of the plates of one capacitor is brought closer to the other plate

A
the voltage on the capacitor whose plates came closer is greater than the voltage on the capacitor whose plates are not moved.

B
the voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor whose plates are not moved.

C
the voltages on the two capacitors remain equal.

D
the applied voltage is divided equally between the two capacitors

Solution

Capacitance is given by, $$C = \dfrac{\epsilon A}{d}$$

where $$A =$$ Area of plates, $$d =$$ distance between the plates

If one plate is brought closed, i.e, $$d$$ decreases, capacitance of that capacitor will increases.

Since, $$C = \dfrac{q}{V}$$

If capacitance of capacitor is increased, Voltage of that capacitor will decreases.
So, Voltage on the capacitor whose plates came closer is smaller than the voltage on the capacitor that is not moved.
Question 4
1 / -0

The capacitor is charged to equilibrium. A dielectric of constant $$K$$ is slowly inserted in the capacitor. Find work done by the external agent in inserting the dielectric.

A
$$\dfrac{1}{2} c\varepsilon^2 (k - 1)$$

B
$$\dfrac{1}{2} c\varepsilon^2 (1 - k)$$

C
$$\dfrac{1}{2} kc\varepsilon^2$$

D
$$-\dfrac{1}{2}kc\varepsilon^2$$

Solution

$$\dfrac{1}{2}c\varepsilon^2; \dfrac{1}{2} kc\varepsilon^2$$

$$\Delta U = \dfrac{1}{2} c\varepsilon^2 (k - 1)$$

$$= -\Delta U$$

$$= \dfrac{1}{2} c\varepsilon^2 (1 - k)$$
Question 5
1 / -0

Two capacitors of caacity $${ C }_{ 1 }$$ and $${ C }_{ 2}$$ are connected in series and potential difference V is applied across it. Then the potential difference across$${ C }_{ 1 }$$ will be

A
$$V\frac { { C }_{ 2 } }{ { C }_{ 1 } } $$

B
$$V\frac { { C }_{ 1 }+{ C }_{ 2 } }{ { C }_{ 1 } } $$

C
$$V\frac { { C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } $$

D
$$V\frac { { C }_{ 1 } }{ { C }_{ 1 }+{ C }_{ 2 } } $$

Solution

$$Given:$$ two capacitors of capacity $$C_1$$ and $$C_2$$ are connected in series and potential difference V is applied across it.
$$Solution:$$ We know that,
$$Q=C_{\text {net }} V$$

$$Q=\left(\dfrac{C_{1} C_{2}}{c_{1}+C_{2}}\right) V$$
$$\therefore$$ $$, $$Voltage across $$C_{1}=\dfrac{Q}{C_{1}}$$
$$=\dfrac{\dfrac{\left(c_{1} c_{2}\right) v}{(\left. c_{1}+c_{2}\right)}}{c_{1}}$$

$$=\dfrac{c_{2}}{c_{1}+c_{2}} v $$

$$So,the$$ $$correct$$ $$option:C$$
Question 6
1 / -0

Two capacitors are first connected in parallel and then in series. If the equivalent capacitances in the two cases are 16 F and 3 F, respectively, then capacitance of each capacitor is

A
$$16F, 3F$$

B
$$12F, 4F$$

C
$$6F, 8F$$

D
$$6F, 16F$$
Question 7
1 / -0

Two points charges $$4 \mu C$$ and $$-2 \mu C$$ are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

A
In the middle of the two charges

B
$$1/3m$$ from $$4\mu C$$

C
$$1/3m$$ from $$-2\mu C$$

D
Nowhere the potential is zero

Solution

Let potential is zero at a distance $$x$$ from $$4\mu c$$
$$ \Rightarrow \dfrac{{k\left( 4 \right)}}{x} + \dfrac{{k\left( { - 2} \right)}}{{\left( {1 - x} \right)}} = 0$$
$$ \Rightarrow \frac{4}{x} = \dfrac{2}{{\left( {1 - x} \right)}}$$
$$ \Rightarrow 2 - 2x = x$$
$$ \Rightarrow 2 = 3x$$
$$ \Rightarrow x = \dfrac{2}{3}$$
$$\therefore $$ from the distance is $$1 - \dfrac{2}{3} = \dfrac{1}{3}$$
potential is zero at $$\dfrac{1}{3}m$$ from $$ - 2\mu c.$$
Hence,
option $$(C)$$ is correct answer.
Question 8
1 / -0

In a parallel plate capacitor, the region between the plates is filled by a delectric slab. The capacitor is connected to a cell and the slab is taken out.

A
Some charge is drawn from the cell

B
Some charge is returned to the cell

C
The potential difference across the capacitor is reduced

D
No work is done by an external agent in taking the slab out

Solution

$$\therefore$$ Some charge is returned to the cell .
$$\therefore$$ Option $$B$$ is correct answer.
Question 9
1 / -0

Two identical particles of mass m carry a charge Q each . Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle a large distance with speed v. The closed distance of approach be

A
$$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{mv}}$$

B
$$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{4{Q^2}}}{{m{v^2}}}$$

C
$$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{2{Q^2}}}{{m{v^2}}}$$

D
$$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{3{Q^2}}}{{m{v^2}}}$$

Solution

Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be $$u$$

So using conservation of momentum we get

$$mv=2mu$$ or $$u=\dfrac{v}{2}$$

The initial energy of the system is given as $$\dfrac{1}{2}mv^2$$

And the energy at the minimum distance is given as

$$\dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2+\dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2+\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$$

Equating the two energies we get

$$\dfrac{1}{2}mv^2=\dfrac{1}{4}mv^2+\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$$

or

$$\dfrac{1}{4}mv^2=\dfrac{1}{4\pi\epsilon_o}\dfrac{Q^2}{R}$$

or

$$R=\dfrac{1}{4\pi \epsilon_0}\dfrac{4Q^2}{mv^2}$$
Question 10
1 / -0

An electric circuit requires a total capacitance of $$2 \mu F$$ across potential of $$1000 V$$ . Large number $$1 \mu F$$ capacitances are available each of which would breakdown if the potential is more than $$350 V$$ How many capacitances are required to make the circuit.

A
$$24$$

B
$$20$$

C
$$18$$

D
$$12$$

Solution

$$\begin{array}{l} { C_{ total } }=2\mu F \\ v=1000v \\ Number\, \, of\, \, capacitor\, \, in\, series \\ =\dfrac { { 1000 } }{ { 30 } } \\ \simeq 3\left( { 2.8 } \right) \\ { C_{ series } }=\dfrac { 1 }{ 3 } \\ Now, \\ 2=P\times \dfrac { 1 }{ 3 } \\ P=6\left( { in\, \, parallel } \right) \\ total\, \, number\, \, of\, capaci\tan ce\, \\ =3\times 6 \\ =18 \\ \therefore \, option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$