Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 37

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

On removing the dielectric from a charged condenser, its energy

A
increase

B
remains unchanged

C
decreases

D
none of these

Solution

As capacitance, $$C \propto k$$ where $$k=$$ dielectric constant.
and potential $$V \propto \frac{1}{k}$$,
Energy $$U=\frac{1}{2}CV^2 \Rightarrow U \propto \frac{1}{k}$$
Thus when dielectric present , energy will decrease and when it is removed ,the energy will increase.
Question 2
1 / -0

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

A
6 F

B
5 F

C
2/3 F

D
3/2 F

Solution

Let the resultant capacitor is $$C_{R}$$
For series combination of three capacitors , $$\dfrac{1}{C_R}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2} $$ F
$$\therefore C_R=\dfrac{2}{3}F$$
Question 3
1 / -0

When two condensers of capacitance $$1\mu F$$ and $$2\mu F$$ are connected is series then the effective capacitance will be :

A
$$\dfrac{2}{3}\mu F$$

B
$$\dfrac{3}{2}\mu F$$

C
$$3\mu F$$

D
$$4\mu F$$

Solution

When two condenser are in series , the equivalent capacitance $$C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{1\times2}{1+2}=\dfrac{2}{3} \mu F$$
Question 4
1 / -0

The minimum number of condensers each of capacitance of $$2 \mu F$$, in order to obtain resultant capacitance of $$5\mu F$$ will be :

A
4

B
10

C
5

D
6

Solution

We can obtain an equivalence capacitance of $$5\mu F$$ by connecting minimum 4 capacitance of each $$2\mu F$$ only in the way as shown in the figure.
Therefore, A is corrct option.
Question 5
1 / -0

Three condenser of capacity $$2\mu F, 4\mu F$$ and $$8 \mu F$$ respectively, are first connected in series and then connected in parallel. The ratio of equivalent capacitances in two cases will be :

A
7:3

B
49:4

C
3:7

D
4:49

Solution

For series, $$\dfrac{1}{C_{series}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{7}{8}$$
$$\therefore C_{series}=\dfrac{8}{7} \mu F$$
For parallel, $$C_{parallel}=C_1+C_2+C_3=2+4+8=14 \mu F$$
$$\dfrac{C_{series}}{C_{parallel}}=\dfrac{8}{7\times 14}=\dfrac{4}{49}$$
Question 6
1 / -0

What will be area of pieces of paper in order to make a paper condenser of capacitance $$0.04 \mu F$$, if the dielectric constant of paper is $$2.5$$ and its thickness is $$0.025 mm$$ ?

A
$$1m^2$$

B
$$2\times 10^{-3} m^2$$

C
$$4.51\times10^{3} m^2$$

D
$$10^{-3} m^2$$

Solution

For paper condenser we will use the expression of capacitance of parallel plate condenser with dielectric.
$$C=\dfrac{Ak\epsilon_0}{d}$$
$$A=\dfrac{Cd}{k\epsilon_0}=\dfrac{0.04\times 10^{-6}\times 0.025\times 10^{-3}}{2.5\times 8.854\times 10^{-12}}=4.51\times 10^{3} m^2$$
Question 7
1 / -0

When dielectric medium of constant k is filled between the plates of a charged parallel-plate condenser, then the energy stored becomes, as compared to its previous value,

A
$$K^{-3} times$$

B
$$K^{-2} times$$

C
$$K^{-1} times$$

D
$$K times$$

Solution

As no cell is connected to the capacitor so the charge (Q) remains constant.
If the capacitance of capacitor before inserting the dielectric is $$C$$ then capacitance after inserting dielectric becomes $$KC$$
Before inserting dielectric the energy is $$U=\dfrac{Q^2}{2C}$$
After inserting dielectric the energy is $$U'=\dfrac{Q^2}{2KC}=UK^{-1}$$
Question 8
1 / -0

A capacitor of capacitance C is connected to battery of emf $$V_0$$. Without removing the battery, a dielectric of strength $$\varepsilon _r$$ is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :

A
$$CV_0$$

B
$$\varepsilon _rCV_0$$

C
$$\dfrac{CV_0}{\varepsilon _r}$$

D
none of these

Solution

As the battery is not removed so the potential remains constant and it is equal to $$V_0$$.
Due to insert the dielectric the capacitance $$C$$ will become $$C'=\varepsilon_r C $$
Thus the charge on the capacitor $$Q=C'V_0=\varepsilon_r CV_0$$
Question 9
1 / -0

The capacitance of a parallel plate capacitor in air is $$2\ \mu F$$. If a dielectric medium is placed between the plates then the potential difference reduces to $$\dfrac{1}{6}$$ of the original value. The dielectric constant of the medium is :

A
$$6$$

B
$$3$$

C
$$2.2$$

D
$$4.4$$

Solution

As no battery is connected to the capacitor, the charge remains constant.
Before inserting the dielectric, the charge is $$Q=CV$$
After inserting the dielectric, the capacitance become $$C'=KC$$ and hence charge $$Q'=\dfrac{C'V}{6}=\dfrac{KCV}{6}$$
As the charge is constant , $$Q=Q'$$
$$ \Rightarrow CV=\dfrac{KCV}{6}$$
$$\therefore K=6$$
Question 10
1 / -0

The capacitance of parallel-plate capacitor is $$4\mu F$$. If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :

A
$$1/64 \mu F$$

B
$$0.25 \mu F$$

C
$$64 \mu F$$

D
$$40 \mu F$$

Solution

If the capacitance of the parallel plate capacitor without dielectric is $$C$$, the capacitance of it with dielectric will be $$C'=kC=16\times 4=64 \mu F$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electrostatic Potential and Capacitance Test - 37

Result

Electrostatic Potential and Capacitance Test - 37

Score

-

Rank

-

SHARING IS CARING

Question 1

increase

remains unchanged

decreases

none of these

Solution

Question 2

6 F

5 F

2/3 F

3/2 F

Solution

Question 3

$$\dfrac{2}{3}\mu F$$

$$\dfrac{3}{2}\mu F$$

$$3\mu F$$

$$4\mu F$$

Solution

Question 4

4

10

5

6

Solution

Question 5

7:3

49:4

3:7

4:49

Solution

Question 6

$$1m^2$$

$$2\times 10^{-3} m^2$$

$$4.51\times10^{3} m^2$$

$$10^{-3} m^2$$

Solution

Question 7

$$K^{-3} times$$

$$K^{-2} times$$

$$K^{-1} times$$

$$K times$$

Solution

Question 8

$$CV_0$$

$$\varepsilon _rCV_0$$

$$\dfrac{CV_0}{\varepsilon _r}$$

none of these

Solution

Question 9

$$6$$

$$3$$

$$2.2$$

$$4.4$$

Solution

Question 10

$$1/64 \mu F$$

$$0.25 \mu F$$

$$64 \mu F$$

$$40 \mu F$$

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.