Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The two capacitors $$2\mu F$$ and $$6\mu F$$ are put in series, the effective capacity of the system in $$\mu F$$ is:

A
$$8$$

B
$$2$$

C
$$\dfrac{3}{2}$$

D
$$\dfrac{2}{3}$$

Solution

$$2\mu F, 6\mu F$$ are in series

$$C_{eq}=\dfrac {C_1C_2}{C_1+C_2}$$

$$=\dfrac {2\times 6}{(2+6)}=\dfrac {12}{8}=\dfrac {3}{2}\mu F$$
Question 2
1 / -0

Electrical potential at the centre of a charged conductor is:

A
zero

B
twice as that on the surface

C
half of that on the surface

D
same as that on the surface

Solution

Inside a charged conductor there is no electric field that is $$E=0$$.

And, we know that $$\dfrac{dV}{dr}=-E=0$$

$$\implies V=constant.$$

Hence, electric potential at center of charged conductor is same as that on surface and at all points in conductor.

Answer_(D)
Question 3
1 / -0

In a cathode ray oscillograph, the focussing of beam on the screen is achieved by

A
Convex lenses

B
Magnetic field

C
Electric potential

D
All of these

Solution
Question 4
1 / -0

When two capacitors of capacities $$3\mu F$$ and $$6\mu F$$ are connected in series and connected to $$120 V,$$ the potential difference across $$3\mu F$$ is:

A
$$40V$$

B
$$60V$$

C
$$80V$$

D
$$180V$$

Solution

$$C=\dfrac {3\times 6}{9}=2\mu F$$
$$V=120 V$$
$$Q=CV=120\times 2=240\mu C$$
$$V_3=\dfrac {Q}{3\mu F}=\dfrac {240\mu C}{2\mu F}=80 V$$
Question 5
1 / -0

If $$C_1=3pF$$ and $$C_2=2pF$$, calculate the equivalent capacitance of the network shown in the figure between points A and B?

A
1 pF

B
2 pF

C
3 pF

D
4 pF

Solution

The given network shown in figure is equivalent to the arrangement shown in figure, which is further equivalent to arrangement shown in figure.
Therefore,
$$\dfrac {1}{C'}=\dfrac {1}{C_1}+\dfrac {1}{C_2}+\dfrac {1}{C_1}$$
$$=\dfrac {1}{3}+\dfrac {1}{2}+\dfrac {1}{3}=\dfrac {7}{6}$$ or $$C'=\dfrac {6}{7}pF$$
Here, C" is capacitance of the parallel combination of C' and $$C_2.$$ Therefore,
$$C"=C'+C_2=\dfrac {6}{7}+2=\dfrac {20}{7}pF$$
If $$C_{equi}$$ is capacitance of the series combination of $$C_1, C" and C_1$$, then
$$\dfrac {1}{C_{equi}}=\dfrac {1}{C_1}+\dfrac {1}{C"}+\dfrac {1}{C_1}=\dfrac {1}{3}+\dfrac {7}{20}+\dfrac {1}{3}$$
$$=\dfrac {61}{60}$$
or $$C_{equi}=\dfrac {60}{61}\approx 1pF$$
Question 6
1 / -0

Find the charge appearing on capacitor of capacitance $$4\;\mu F$$ :

A
$$32\;\mu C$$

B
$$96\;\mu C$$

C
$$12\;\mu C$$

D
$$4\;\mu C$$

Solution

As the two capacitances are in series combination, so equivalent capacitance,$$C_{equ}=\dfrac{8\times4}{8+4}=\dfrac{32}{12}=\dfrac{8}{3} \mu F$$
So, charge on $$4\ \mu F, Q=C_{equ}\times V=\dfrac{8}{3}\times12=32 \ \mu C$$
Question 7
1 / -0

Condensers of capacities $$2\mu F$$ and $$3\mu F$$ are connected in series and a condenser of capacity $$1\mu F$$ is connected in parallel with them. The resultant capacity is:

A
$$0.5\mu F$$

B
$$1.1\mu F$$

C
$$22\mu F$$

D
$$2.2\mu F$$

Solution

$$\dfrac{6}{5}+1=\dfrac{11}{5}=2.2\mu F$$.
Question 8
1 / -0

What is the equivalent capacitance of the system of capacitors between $$A \mbox{ & } B$$?

A
$$\displaystyle\frac{7}{6}C$$

B
$$1.6 C$$

C
$$C$$

D
None

Solution

The capacitors in the right hand square are in series. So they can be replaced by an equivalent capacitor of capacitance of:
$$1/(1/C+1/C) = C/2$$
The next step is to replace C and C/2 by a capacitor having an equivalent capacitance of C and C/2 in parallel.
$$C+C/2=3C/2$$
Then, C and 3C/2 are in series. So, they can be replaced by a capacitor having an equivalent capacitance of
$$1/(1/C+2/3C)=3C/5$$
Then since in parallel they can be replaced by an equivalent capacitance of $$C+3C/5=8C/5 = 1.6C$$
So the equivalent capacitance is 1.6C.
Question 9
1 / -0

If $$4\times 10^{20}eV$$ of energy is required to move a charge of 0.25 coulomb between two points, the p.d between them is:

A
256 V

B
512 V

C
123 V

D
215 V

Solution

$$\displaystyle V=\frac {W}{Q}=\frac {4\times 10^{20}\times 1.6\times 10^{-19}J}{0.25C}$$
$$=256 V$$
Question 10
1 / -0

A capacitor stores $$60\space\mu C$$ charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of $$120\space\mu C$$ flows through the battery. The dielectric constant of the material inserted is :

A
$$1$$

B
$$2$$

C
$$3$$

D
none

Solution

The capacitance before filled with dielectric is $$C=\dfrac{A\epsilon_0}{d}$$
The capacitance after filled with dielectric is $$C'=\dfrac{Ak\epsilon_0}{d}=kC$$
Initial charge, $$Q=60 \mu C$$ and total charge after filled with dielectric $$Q'=120+60=180 \mu C$$
Here, $$Q=CV$$ and $$Q'=C'V=kCV$$
$$\dfrac{Q'}{Q}=\dfrac{kCV}{CV}$$
$$k=\dfrac{Q'}{Q}=\dfrac{180}{60}=3$$