Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The radius of a nucleus of an atom $$(Z=50)$$ is $$9\times 10^{-10}m$$ then potential on its surface, will be:

A
$$80V$$

B
$$8kV$$

C
$$9V$$

D
$$9kV$$

Solution

$$R=9\times 10^{-10}m$$
$$q=charge=ze$$
$$q=50\times 1.6\times 10^{-19}C$$
$$V=potential =\dfrac{kq}{R}$$
$$V=\dfrac{9\times 10^9\times 50\times 1.6\times 10^{-19}}{9\times 10^{-10}}$$
$$V=80$$Volt
Question 2
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The distance between at the plates of a charged plate capacitor disconnected from the battery is 5 cm and the intensity of the field in it is E = 300 V/cm. An uncharged metal bar 1 cm thick is introduced into the capacitor parallel between its plates. If the battery provided 6 uC charge, find final capacitance.

A
4 $$\mu f$$

B
5 $$\mu f$$

C
6 $$\mu f$$

D
8 $$\mu f$$

Solution

Before inserting plate:-
Charge gained =$$6\mu c$$ and electric field $$=300 /cm$$
the voltage developed across the plates,
$$V=E.d =300\times 5=1500V$$
Thus capacitance $$c=\dfrac{Q}{V}=\dfrac{6\times 10^{-6}}{1500}=4\times 10^{-9}=4nF$$
now the mental plate in not dieretric i.e when inserted it would not generate the reverse field and thus the capacitance wouldn't be affected
final capacitance $$4\times 10^{-9}=4nF$$
Question 3
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A circuit
has a section AB as shown in figure with E=$${\text{10}}\;\;{\text{V,}}\;{{\text{C}}_{{\text{1}}\;}}{\text{ = }}$$ 1.0 $$\mu F$$$${{\text{C}}_{\text{2}}}\;{\text{ = }}\;{\text{2}}{\text{.0}}$$ $$\mu F$$and the potential difference $${{\text{V}}_{\text{A}}}\;{\text{ - }}{{\text{V}}_{\text{B}}}{\text{ = 5}}\;{\text{V}}$$ The
voltage across $${{\text{C}}_{\text{1}}}$$ is :

A
Zero

B
5 V

C
10 V

D
15 V

Solution

The saturation period, there is no voltage gain across the equation.
Question 4
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During charging and discharging of a capacitor:-

A
Current flows in the circuit, which is constant during charging or discharging duration

B
No current flows in the circuit

C
Current flows in the circuit and is varying with time

D
During chargung current is constant but while discharging current is variable

Solution

During charging of capacitor, the charge on capacitor increases and hence voltage across capacitor increases as, $$Q = CV$$. Due to increase in voltage, the flow of current increases.
Similarly, during dicharging of capacitor, the charge on capacitor decreases and hence voltage across capacitor decreases. Due to decrease in voltage, the flow of current also decreases.

So current flows in circuits and is varying with time.
Question 5
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In the given circuit the effective capacity between A and B is

A
$$20 { \mu } F$$

B
$$5 { \mu } F$$

C
$$30 { \mu } F$$

D
$$10 { \mu } F$$

Solution

Let $$\sim$$ denote series combination,

And, $$||$$ denote parallel combination,

Formulating the circuit,

$$((((12\sim 12)||4)\sim10) || (10\sim 10) ) \sim 10$$

$$=((10\sim 10) || (10\sim 10) ) \sim 10 = (5 || 5) \sim 10$$

$$=10\sim 10=5\mu F$$

Option $$\textbf B$$ is the correct answer
Question 6
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Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is:

A
Q/2V

B
2Q/V

C
Q/V

D
None of these

Solution

In series connection charge on each capacitor would be constant also equivalent capacitance in series $$c'=\dfrac{C}{2}$$ [ following $$\dfrac{1}{c'}=\dfrac{1}{c_1}+\dfrac{1}{c_2}$$] and voltage $$V$$ is applied across it so,from capacitive law,$$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$$
Question 7
1 / -0

Choose the incorrect relation among the following:

A
Ampere = $$ \frac{coulomb}{second} $$

B
Volt = $$ \frac{coulomb}{joule} $$

C
ohm = $$ \frac{Volt}{Ampere} $$

D
$$ 1 coulomb = 6.25 \times 10^{18} electron charge $$

Solution

Ampere = $$SI$$ unit of current $$=\dfrac{Charge (Coulomb)}{Time (Second)}$$
Volt = $$SI$$ unit of Potential $$=\dfrac{Work (Joule)}{Charge (Coulomb)}$$
Ohm = $$SI$$ unit of Resistance $$=\dfrac{Potential (Volt)}{Current (Ampere)}$$
$$1\ Coulomb = (1.6\times 10^{-19})^{-1}$$ electron charge
$$=6.25\times 10^{18}$$ electron charge
Option $$B$$ is incorrect.
Question 8
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Three identical capacitors are connected in series. The capacitance of combination is $$C$$ and break-down voltage is $$V$$. The net capacitance and break down voltage when these are connected in parallel combination will be:

A
$$\dfrac{C}{3}, 3V$$

B
$$9C, \dfrac{V}{3}$$

C
$$\dfrac{C}{3}. V$$

D
$$3C, \dfrac{V}{9}$$

Solution

In series
$$\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$$
here $$C_1=C_2=C_3=C$$

Then net capacitance is
$$C'=\dfrac{C}{3}$$ and V is given

So charge $$q=CV$$ putting values

$$q=\dfrac{VC}{3}$$

Now when capacitors are connected in parallel
$$C"=C+C+C=3C$$

and charge will remain same
$$V'=\dfrac{q}{C"}$$

$$V'=\dfrac{V}{9}$$
Question 9
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You are given 32 capacitors of $$4\mu F$$ capacitance each. How do you connect all of them so that the effective capacitance becomes $$8 \mu F$$?

A
$$4$$ capacitors in series and $$8$$ such groups in parallel.

B
$$2$$ capacitors in series and $$16$$ such groups in parallel.

C
$$8$$ capacitors in series and $$4$$ such groups in parallel.

D
All of them in series.
Question 10
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What is the electric potential at a distance $$'x'$$ from the centre, inside a conducting sphere having a charge $$Q$$ and radius $$R$$?

A
$$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{R}$$

B
$$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{x}$$

C
$$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {QX}{R^{2}}$$

D
Zero

Solution

This is a direct formula for JEE or medical students