Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A charge of 5 C is moved between two points in an electric field and 20 J of work was done to do so. Calculate the potential difference between the two points

A
4 V

B
2 V

C
3 V

D
6 V

Solution

Given:
charge to be moved in an electric field$$=5C=q$$
work done$$=20J$$

we have to find potential difference b/w 2 point.
here, we use the formula that

$$work done =(charge\times potential\quad difference)$$ b/w 2 point.
so, potential difference b//w 2 point
$$V=\cfrac { work\quad done }{ charge } \\ =\cfrac { 20 }{ 5 } =4V$$.
Question 2
1 / -0

Seven capacitors, each of capacitance $$2\mu F$$ are to be connected to obtain a capacitance of $$10/11\mu F$$. Which of the following combinations is possible?

A
$$5$$ in parallel $$2$$ in series

B
$$4$$ in parallel $$3$$ in series

C
$$3$$ in parallel $$4$$ in series

D
$$2$$ in parallel $$5$$ in series

Solution

When 5 in parallel 2 in series, $$C_{eq}=\dfrac{(5\times 2)(2/2)}{(5\times 2)+(2/2)}=\dfrac{10}{10+11}=\dfrac{10}{11} \mu F $$
When 4 in parallel 3 in series , $$C_{eq}=\dfrac{(4\times 2)(2/3)}{(4\times 2)+(2/3)}=\dfrac{16/3}{26/3}=\dfrac{8}{13} \mu F$$
When 3 in parallel 4 in series , $$C_{eq}=\dfrac{(3\times 2)(2/4)}{(3\times 2)+(2/4)}=\dfrac{3}{13/2}=\dfrac{6}{13} \mu F$$
When 2 in parallel 5 in series , $$C_{eq}=\dfrac{(2\times 2)(2/5)}{(2\times 2)+(2/5)}=\dfrac{8/5}{22/5}=\dfrac{4}{11} \mu F$$
Question 3
1 / -0

Two identical capacitors are joined in parallel and charged to a potential V. Then these are separated and connected in series with the positive plate of one connected to the negative ofthe other. Then the

A
the potential difference between the plates becomes 2V

B
the energy stored in the system increases

C
the charges on the free plates are enhanced

D
the charges on the plates connected together are destroyed

Solution

When they are connected in parallel, each capacitor will have a potential V. When they are joined in series having each of potential V, then total potential across the two extreme plates of the capacitor = V + V = 2V.
Question 4
1 / -0

Three capacitors each of 1 microfarad are connected as shown below, The capacitance between the points $$A$$ and $$B$$ is :

A
$$3 \mu F$$

B
$$1 \mu F$$

C
$${2}/{3} \mu F$$

D
$${1}/{3} \mu F$$

Solution

For parallel connection, equivalent capacitance of $$C_1$$ and $$C_2$$ $$C_p = C_1+C_2 = 2\mu F$$
$$\therefore$$ Equivalent capacitance between A and B $$\dfrac{1}{C_{eq}} =\dfrac{1}{C_p}+\dfrac{1}{C_3}$$
$$\therefore$$ $$\dfrac{1}{C_{eq}} = \dfrac{1}{2}+\dfrac{1}{1}$$ $$\implies C_{eq} = \dfrac{2}{3}\mu F$$
Question 5
1 / -0

$$ECG$$ is a method of diagnosis used to check the functioning of heart, it is based on the principle that there is

A
A pattern of magnetic field associated with the functioning of heart

B
An electrical impulse associated with the functioning of organs and tissues of human body

C
A varying electrical field associated. with organs and tissues of human body

D
A varying magnetic field associated with organs and tissues of human body

Solution

The heart generates by itself an electrical activity which is transmitted through all the organ and produces its contraction. The Electrocardiogram is no other than the graphical representation of this electrical activity.
Question 6
1 / -0

1 V equals :

A
1 J

B
1 JC$$^{-1}$$

C
1 CJ$$^{-1}$$

D
1 JC

Solution

1 volt is equal to 1 joule of electric potential energy per (divided by) 1 coulomb of charge.

V= J/C = joule/coulomb

Option B is correct.
Question 7
1 / -0

What is the resultant capacitance between points A and B?

A
51 $$\mu$$F

B
102 $$\mu$$F

C
204 $$\mu$$F

D
408 $$\mu$$F

Solution

Let $$C1=200\mu F \ $$ $$C2=20\mu F \ $$ and $$C3=5\mu F \ $$
In the given circuit $$\left ( C2\ series \ C3 \right ) \ || \ C1$$
$$\therefore C_{eq} =\left ( \dfrac{200 \times5}{200+5} \right )+200 \\$$
$$\therefore C_{eq} =\dfrac{1000+200 \times205}{205}$$
$$\therefore C_{eq} = 204.87 \mu F$$
Therefore the resultant capacitance between A and B is $$C ) \ 204 \mu F$$
Question 8
1 / -0

The electric potential at the surface of an atomic nucleus (Z = 50) of radius $$9.0 \times 10^{-13} cm$$ is:

A
$$ 9\times 10^5$$ volt

B
$$ 8 \times 10^6$$ volt

C
80 volt

D
9 volt

Solution

Electric potential at the surface of an atomic nucleus,
$$V = k. \displaystyle \dfrac{Ze}{r}$$
$$=\displaystyle \frac{9\times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} = 8 \times 10^6 V$$
Question 9
1 / -0

In the circuit shown in the figure, the potential difference across the $$4.5 \mu F$$ capacitor is:

A
$$\displaystyle \frac{8}{3}$$ volt

B
4 volt

C
6 volt

D
8 volt

Solution

Total capacitance, $$\displaystyle \frac{1}{C} =\frac{1}{4.5} +\frac{1}{9}=\frac{1}{3}$$
or $$C = 3 \mu F$$
Charge, Q = CV = 3 $$\mu $$ F $$\times $$ 12V = 36 $$\mu$$ C
Potential difference across $$4.5 \mu F$$ capacitor

$$=\displaystyle \frac{36 \mu C}{4.5 \mu F} =8V$$
Question 10
1 / -0

Minimum numbers of $$\displaystyle 8\mu F$$ and 250 V capacitors are used to make a combination of $$\displaystyle 16\mu F$$ and 1000 V are:

A
4

B
32

C
8

D
3

Solution

To create 1000 Y, we need to combine 4 capacitors
in series. Total capacity becomes $$\dfrac{8 \mu F}{4} = 2 \mu F$$.In order to obtain capacity of $$16 \mu F$$, 8-rows of this combination will be needed in parallel.
Total capacity $$= 2 \mu F \times 8 = 16 \mu F$$
Total number of capacitor $$ = 4 \times 8 = 32$$

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Electrostatic Potential and Capacitance Test - 41

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Electrostatic Potential and Capacitance Test - 41

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Question 1

4 V

2 V

3 V

6 V

Solution

Question 2

$$5$$ in parallel $$2$$ in series

$$4$$ in parallel $$3$$ in series

$$3$$ in parallel $$4$$ in series

$$2$$ in parallel $$5$$ in series

Solution

Question 3

the potential difference between the plates becomes 2V

the energy stored in the system increases

the charges on the free plates are enhanced

the charges on the plates connected together are destroyed

Solution

Question 4

$$3 \mu F$$

$$1 \mu F$$

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

$${1}/{3} \mu F$$

Solution

Question 5

A pattern of magnetic field associated with the functioning of heart

An electrical impulse associated with the functioning of organs and tissues of human body

A varying electrical field associated. with organs and tissues of human body

A varying magnetic field associated with organs and tissues of human body

Solution

Question 6

1 J

1 JC$$^{-1}$$

1 CJ$$^{-1}$$

1 JC

Solution

Question 7

51 $$\mu$$F

102 $$\mu$$F

204 $$\mu$$F

408 $$\mu$$F

Solution

Question 8

$$ 9\times 10^5$$ volt

$$ 8 \times 10^6$$ volt

80 volt

9 volt

Solution

Question 9

$$\displaystyle \frac{8}{3}$$ volt

4 volt

6 volt

8 volt

Solution

Question 10

4

32

8

3

Solution

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