Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 42

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

Directions For Questions

These questions consist of two statements, each printed as assertion and reason. While answering these question you are required to choose any one of the following five responses.

...view full instructions

Assertion: If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes 6 times.
Reason : Capacity of the capacitor does not depend upon the nature of the material.

A
If both assertion and reason are true but the reason is the correct explanation of assertion.

B
If both assertion and reason are true but the reason is not the correct explanation of assertion.

C
If assertion is true but reason is false.

D
If both the assertion and reason are false.

E
If reason is true but assertion is false.

Solution

We know that capacity of capacitor is directly proportional to dielectric constant and inversely proportional to distance. So, the net effect of making distance halved & making dielectric constant three times will be capacity becoming six times.
As nature of the material (dielectric constant) is a factor influencing the capacity, therefore, reason is false.
Question 2
1 / -0

Which of the following is not a unit for electric potential?

A
Volt

B
Joule/coulomb

C
Erg/stat coulomb

D
None of these

Solution

Volt is SI unit of electric potential.
Also, $$V=\dfrac{Work}{Charge}=\dfrac{J}{Coulomb}$$
In C.G.S, Volt $$=\dfrac{Erg}{Stat\;Coulomb}$$

So, a, b, c are units of potential.
Question 3
1 / -0

A parallel plate air capacitor has a capacitance $$C$$.When it is half filled with a dielectric of dielectric constant $$5$$, the percentage increase in the capacitance will be :

A
$$400$$ %

B
$$66.6$$ %

C
$$33.3$$$ %

D
$$200$$ %

Solution

Initial capacitance $$= \dfrac{\epsilon_oA}{d}$$
When it is half filIed by a dielectric of dielectric constant K, then
$$C_1 = \dfrac{K\epsilon_oA}{d/2} = 2K \dfrac{\epsilon_oA}{d}$$
and $$C_2 = \dfrac{\epsilon_oA}{d/2} = \dfrac{2\epsilon_oA}{d}$$

$$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{d}{2\epsilon_oA}(\dfrac{1}{K} + 1) = \dfrac{d}{2\epsilon_oA}(\dfrac{1}{5}+1) = \dfrac{3}{5} \dfrac{d}{\epsilon_oA}$$
$$C = \dfrac{5}{3} \dfrac{\epsilon_oA}{d}$$
Hence % increase in capacitance
$$\bigg (\dfrac{\dfrac{5\epsilon_oA}{d}-\dfrac{\epsilon_oA}{d}}{\dfrac{\epsilon_oA}{d}}\bigg ) \times 100 = \dfrac{2}{3} \times 100 = 66.6$$%
Question 4
1 / -0

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in :

A
increase in the potential difference across the plates and reduction in stored energy but no change in the charge on the plates

B
decrease in the potential difference across the plates and reduction in the stored energy but no change in the charge on the plates

C
reduction of charge on the plates and Increase of potential difference across the plates

D
increase in stored energy but no change in potential difference across the plates

Solution

If a dielectric slab of dielectric constant K is filled in between the plates of a capacitor after charging the capacitor (i.e., after removing the connection of battery with the plates of capacitor) the potential difference between the plates reduces to $$\dfrac{1}{K}$$ times and the potential energy of capacitor reduces to $$\dfrac{1}{K}$$ . times but there is no change in the charge on the plates.
Question 5
1 / -0

The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is $$6\ \mu F$$. The capacity of each capacitor is :

A
$$2\ \mu F$$

B
$$4\ \mu F$$

C
$$8\ \mu F$$

D
$$16\ \mu F$$

Solution

Let the capacitance of each capacitor be $$C$$
For series combination : $$\dfrac{1}{C_s} = \dfrac{1}{C} + \dfrac{1}{C}$$
$$\implies C_s = \dfrac{C}{2}$$
For parallel combination : $$C_p = C + C$$ $$\implies C_p = 2C$$
Given : $$C_p - C_s = 6\ \mu F$$
$$\Rightarrow $$ $$2C - \dfrac{C}{2} = 6\ \mu F$$
$$\implies C = 4\ \mu F$$
Question 6
1 / -0

An air filled parallel plate condenser has a capacity of $$2pF$$. The separation of the plates is doubled and the inter-space between the plates is filled with wax. If the capacity is increased to $$6$$pF, the dielectric constant of wax is.

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

$$C_{old}=\cfrac{\varepsilon_oA}{d}=2pF\longrightarrow(i)$$
$$C_{new}=\cfrac{K\varepsilon_oA}{2d}=6pF\longrightarrow(ii)$$
By Equating $$(i)$$ and $$(ii)$$
$$\cfrac{K}{2}\left(2\right)pF=6pF$$
$$\Rightarrow K=6$$
Question 7
1 / -0

Three capacitors of capacitances $$1 \mu F$$, $$2 \mu F$$ and $$4 \mu F$$ are connected first in a series combination, and then in a parallel combination. The ratio of their equivalent capacitances will be:

A
$$2 : 49$$

B
$$49 : 2$$

C
$$4 : 49$$

D
$$49 : 4$$

Solution

Equivalent capacitance in series is reciprocal of sum of reciprocals of individual capacitances.
$$ \Rightarrow \dfrac{1}{C_{S}}=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{4}$$
So, the series equivalent is $$\dfrac{4}{7}{\mu}F$$.
In parallel, equivalent capacitance is the sum of individual capacitances.
$$\Rightarrow C_{P}=1+2+4$$
So, the parallel equivalent is $$7{\mu}F$$.
Thus, the ratio is $$\dfrac{4}{49}$$.
Question 8
1 / -0

When a dielectric slab is introduced between the two plates of condenser then its capacity ___________.

A
remains constant

B
increases

C
decreases

D
may increase or decrease depending on the material of dielectric slab

Solution

As the dielectric slab is introduced there is some charge distribution in the slab and because of this the electric field between the two plates is decreased, due to which the capacitor can hold more charge. Thus, the capacity to hold charge of the capacitor is increased.
Question 9
1 / -0

A parallel plate condenser with oil (dielectric constant $$2$$) between the plates has capacitance $$C$$. If the oil is removed, the capacitance of capacitor becomes :

A
$$\sqrt {2C}$$

B
$$2C$$

C
$$\cfrac{C}{\sqrt {2}}$$

D
$$\cfrac{C}{2}$$

Solution

The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is
$$C=\cfrac { K{ \varepsilon }_{ 0 }A }{ d } ........(i)$$
Where $${ \varepsilon }_{ 0 }=$$ electric permitivity of free space
$$K=$$dielectric constant
$$A=$$ area of each plate of capacitor
$$d=$$distance between two plates
When dielectric (oil) is removed, so capacitance
$${C}_{0}=\cfrac { { \varepsilon }_{ 0 }A }{ d } ............(ii)$$
Comparing Eqs. $$(i)$$ and $$(ii)$$ we get
$$C=K{C}_{0}$$
$$\Rightarrow$$ $${C}_{0}=\cfrac{C}{K}=\cfrac{C}{2}$$ ($$\because K=2$$)
Question 10
1 / -0

If dielectric is inserted in charged capacitor (battery removed ), then quantity that remains constant is.

A
Capacitance

B
Potential

C
Intensity

D
Charge

Solution

Variation of different variables $$(Q, C, V, E$$ and $$U)$$ of parallel plate capacitor when dielectric $$(K)$$ is introduced when battery is removed is
$$C'=KC$$ $$E'=E/K$$
$$Q'=Q$$ $$U'=U/K$$
$$V'=V/K$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electrostatic Potential and Capacitance Test - 42

Result

Electrostatic Potential and Capacitance Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

If both assertion and reason are true but the reason is the correct explanation of assertion.

If both assertion and reason are true but the reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If both the assertion and reason are false.

If reason is true but assertion is false.

Solution

Question 2

Volt

Joule/coulomb

Erg/stat coulomb

None of these

Solution

Question 3

$$400$$ %

$$66.6$$ %

$$33.3$$$ %

$$200$$ %

Solution

Question 4

increase in the potential difference across the plates and reduction in stored energy but no change in the charge on the plates

decrease in the potential difference across the plates and reduction in the stored energy but no change in the charge on the plates

reduction of charge on the plates and Increase of potential difference across the plates

increase in stored energy but no change in potential difference across the plates

Solution

Question 5

$$2\ \mu F$$

$$4\ \mu F$$

$$8\ \mu F$$

$$16\ \mu F$$

Solution

Question 6

$$2$$

$$3$$

$$4$$

$$6$$

Solution

Question 7

$$2 : 49$$

$$49 : 2$$

$$4 : 49$$

$$49 : 4$$

Solution

Question 8

remains constant

increases

decreases

may increase or decrease depending on the material of dielectric slab

Solution

Question 9

$$\sqrt {2C}$$

$$2C$$

$$\cfrac{C}{\sqrt {2}}$$

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

Solution

Question 10

Capacitance

Potential

Intensity

Charge

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.