Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 52

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Question 1
1 / -0

A series combination of $${ n }_{ 1 }$$ capacitors, each of value $${ C }_{ 1 }$$ , is charged by a source of potential different 4 V. When another parallel combination of $${ n }_{ 2 }$$ capacitors, each of value $${ C }_{ 2 }$$, is charged y a source of potential difference V, it has the same(total) energy stored in it, as the first combination has. The value $${ C }_{ 2 }$$, in terms of $${ C }_{ 1 }$$, is then;

A
$$\dfrac { { 2C }_{ 1 } }{ { n }_{ 1 }{ n }_{ 2 } }$$

B
$$16\dfrac { { n }_{ 2 } }{ { n }_{ 1 } } { C }_{ 1 }$$

C
$$2\dfrac { { n }_{ 2 } }{ { n }_{ 1 } } { C }_{ 1 }$$

D
$$\dfrac { { 16 }_{ C1 } }{ { n2 }_{ n1 } } $$

Solution

1st case:-
Energy $$=\cfrac{1}{2}CV^2\\ =\cfrac{1}{2}(\cfrac{C_1}{n_1})(4V)^2\\=\cfrac{8C_1V^2}{n_1}$$
2nd case:-
Energy$$=\cfrac{1}{2}cV^2=\cfrac{1}{2}(n_2C_2)V^2$$
According to question-
$$\cfrac{8C_1V^2}{n_1}=\cfrac{n_2C_2V^2}{2}\\ \Rightarrow \cfrac{8C_1}{n_1}=\cfrac{n_2C_2}{2}\\ \Rightarrow \cfrac{16C_1}{n_1n_2}=C_2$$
Question 2
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A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then

A
capacitance will increase.

B
energy stored will decrease.

C
electric field will increase.

D
Voltage will decrease.

Solution

When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored $$\left( u=\cfrac { { q }^{ 2 } }{ 2C } \right) $$ will increase
Potential $$\left( V=\cfrac { q }{ C } \right) $$ will also increase
Electric field $$\left( E=\cfrac { V }{ D } \right) $$ will increase
Question 3
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A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

A
increases

B
decreases

C
does not change

D
becomes zero

Solution

Capacitance $$C=\varepsilon\dfrac{A}{d}$$ , $$A-area\ of\ the\ capacitor\ ; d-distance\ between\ the\ plates$$
For a charged capacitor $$Q=CV$$
As the distance between the plates of the capacitor is increased the capacitance $$C$$ decreases $$\ \ \ (\because C \propto \dfrac{1}{d})$$
As capacitance decreases the potential difference between the plates $$V \ increases$$
Question 4
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Three capacitors each of capacity 4 $$\mu F$$ are to be connected in such a way that the effective capacitance is 6 $$\mu F$$. This can be done by:

A
connecting them in series

B
connecting them is parallel

C
connecting two in series and one in parallel

D
connecting two in parallel and one in series
Question 5
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A capacitor consists of two metal plates each $$10{\text{ }}cm$$ by $$20{\text{ }}cm;$$ they are separated by a $$2.0{\text{ }}mm$$ thick insulator with dielectric constant $$4.1$$ and dielectric strength 6.0107 V/m. What is the capacitance in $$pF\left( {{{10}^{ - 12}}F} \right)?$$

A
75

B
100

C
240

D
360

Solution

$$C = \cfrac{k\epsilon_0A}{d}$$

$$C =\cfrac{4.1\times 8.85\times 10^{-12}\times 10\times 20\times 10^{-4}}{2\times 10^{-3}}$$

$$C = 361\approx 360 pF$$
Question 6
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Find the equivalent capicitance between points $$A$$ and $$B$$ of the circuit shown, each capacitance $$= C$$.

A
2C

B
3C

C
C

D
4C

Solution

(1) Applying folding symmetry the circuit can be reduced to,
Question 7
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Equipotential surfaces are

A
Surfaces that are perpendicular to gravitational fields

B
Surfaces that have same mass on top of it

C
Surface that have same radius of curvature

D
Surfaces that have same gravitational field on it
Solution
- Equipotential surfaces are surfaces which have the same value of potential throughout their surface.
- For a gravitatiomal field of $$E=\dfrac{GM}{r^2}$$,the potential is $$v=-\dfrac{GM}{r}$$
- So,at a particular 'r',the potential remains constant at a value of $$-\dfrac{GM}{r}$$
- So equipotential surfaces are circles for the filed of $$E=\dfrac{GM}{r^2}$$
- Hence,the field is perpendicular to equipotential surface.
Question 8
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Work done in moving an object through an equipotential surface is

A
Positive

B
Negative

C
Zero

D
Depends on the field direction

Solution

Work done is given difference in potentials. In an equipotential surface, all points will have same potential. Thus work done is zero

The correct option is (c)
Question 9
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An example of an equipotential surface in earth is

A
A line passing through the centre of the earth connecting two points along the diameter

B
A plane that passes through the circular section of the hemisphere of the earth

C
A spherical surface at a distance of 1km from the surface of the earth with its centre at centre of earth

D
A plane on the surface of the earth, which is a tangent to the earth

Solution

The potential is seen to be a constant on a sphere at all points. Hence (c) is the correct option
Question 10
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A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.

A
10

B
15

C
5

D
8

Solution

$$\dfrac{{\varepsilon A}}{d} = \dfrac{{{\varepsilon _0}A}}{{d' - t + \dfrac{r}{{{\varepsilon _r}}}}}$$
or
$$d = d' - t + \dfrac{t}{{{\varepsilon _r}}}$$
$$r - \dfrac{r}{{{\varepsilon _r}}} = d' - d = 2.4\,mm$$
$$r\left( {1 - \dfrac{1}{{{\varepsilon _r}}}} \right) = 2.4$$
$$1 - \dfrac{1}{{{\varepsilon _r}}} = \dfrac{{2.4}}{3}\, \Rightarrow \,{\varepsilon _r} = \dfrac{{30}}{6} = 5$$