Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 60

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Question 1
1 / -0

A parallel plate capacitor consists of two circular plates each of radius 2 cm, separatrd by a distance of 0.1 mm If Voltage across the plates is at the rate of 5$$\times 10^{13}$$ V/s, then the value of displacement current is

A
5.50A

B
5.56$$\times 10^{3}$$A

C
2.28$$\times 10^{4}$$A

D
5.56$$\times 10^{2}$$

Solution

$$A=\pi 0^{2}$$
$$=4\pi \times 10^{-4}$$
$$I_{d}=\in _{0}A \left(\dfrac{dE}{dt}\right)$$
$$=8.85\times 10^{-12}\times 4\pi \times 10^{-4}\times 5\times 10^{13}$$
$$= 555.78\times 10^{-3}A$$
Question 2
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For high frequency a capacitor offer

A
more reactance

B
less reactance

C
zero reactance

D
inifinite reactance

Solution

The relation between the reactance of capacitor and the frequency is given by
$$X_C=\dfrac{1}{\omega C}$$
$$X_C=\dfrac{1}{2\pi f C}$$
$$X_c\propto \dfrac{1}{f}$$
At high frequency, capacitor offer less reactance.
The correct option is B.
Question 3
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In an adjoining figure are shown three capacitors $$\mathrm { C } _ { 1 } , \mathrm { C } _ { 2 }$$ and $$\mathrm { C } _ { 3 }$$ joined to a battery. The correct condition will be:

A
$$Q _ { 1 } = Q _ { 2 } = Q _ { 3 }$$ and $$V _ { 1 } = V _ { 2 } = V _ { 3 } = V$$

B
$$Q _ { 1 } = Q _ { 2 } + Q _ { 3 }$$ and $$V = V _ { 1 } + V _ { 2 }$$

C
$$Q _ { 1 } = Q _ { 2 } + Q _ { 3 }$$ and $$ V = V _ { 1 } + V _ { 2 } + V _ { 3 }$$

D
$$Q _ { 1 } = Q _ { 3 }$$ and $$V _ { 2 } = V _ { 3 }$$

Solution
Question 4
1 / -0

The value of equivalent capacitance of the combination shown in figure, between the points $$P$$ and $$Q$$ is

A
$$3C$$

B
$$2C$$

C
$$C$$

D
$$C/3$$

Solution

A. 3C

the left C and C , capacitors are in parallel to the add up to equivalent of 2C
now, this 2C is in series with the left middle 2C so, it will add up to Ceq of C...
again this C is parallel with middle centre C , adding up to 2C

now again, this 2C is in series with the left upper 2C so, it will add up to Ceq of C...

and finally this C is in parallel with 2C
so, the final Ceq is $$C +2C = 3C$$
Question 5
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If electric intensity $$\overrightarrow { E } $$ is along the X-axis, then the equipotential surfaces are parallel to

A
XOY plane

B
XOZ plane

C
YOZ plane

D
None of these

Solution

Electric intensity $$\left( {\overrightarrow E } \right)$$ is always perpendicular to equipotential surface$$.$$ Since direction of electric intensity is along $$x-$$ axis so equipotential plane will be $$YOZ$$
HJence,
option $$(C)$$ is correct answer.
Question 6
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The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is $$60$$ volts. Then the equivalent capacity between A and B and the charge on $$5\,\mu F$$ capacitance will be respectively

A
$$44\,\mu F;300\mu C$$

B
$$16\,\mu F;150\mu C$$

C
$$15\,\mu F;200\mu C$$

D
$$4\,\mu F;50\mu C$$
Question 7
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The equivalant capacitance of the combination shown in figure below is

A
$$2C$$

B
$$C$$

C
$$\dfrac {1}{2 }C$$

D
$$None$$

Solution

Since, the potential at point A is equal to the potential at point B, no current will flow along the arm AB. Hence, the capacitor on the arm AB will not contribute to the circuit. Also, because the remaining two capacitors are connected in parallel, the net capacitance of the circuit is given by

C_eq=C₁+C₂

C_eq=2C+C=3C

So, equivalent capacitance of above combination is 3C.

So, option (D) is correct.
Question 8
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P, Q and R are three points in a uniform electric field. The electric potential is

A
minimum at R

B
minimum at Q

C
minimum at P

D
Same at all three points

Solution

We know that electric field is directed along decreasing potential.
$$\therefore {V_Q} > {V_R} > {V_P}$$
$$\therefore$$ electric potential is minimum at $$P$$
hence, option $$C$$ is correct.
Question 9
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The ratio of charge densities on the surface of two conducting spheres is 3 : 2. lithe radii of t: the spheres are 4 cm and 8 cm the ratio of the electric potential on the surfaces of the sphere 2 is

A
3 : 4

B
3 : 1

C
1 : 3

D
4 : 9

Solution

Ratio of charge densities $$=3:2$$ spheres are $$4cm$$ and $$8cm$$
spheres ratio $$=4:8=1:2$$
$${ E }_{ 1 }:{ E }_{ 2 }=?$$
$$E$$ $$\alpha$$ $$\dfrac { \lambda }{ r } $$
$$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } \times \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } $$
$$=\dfrac { 3 }{ 2 } \times \dfrac { 2 }{ 1 } $$
$$=3:1$$
$${ E }_{ 2 }:{ E }_{ 1 }=1:3$$
Question 10
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In the circuit shown, two parallel plate capacitors of capacitance $$C$$ and $$2C$$ consisting of plate of identical dimension. They are connected with an ideal battery of terminal voltage $$V$$ as shown in figure $$I$$. Now the capacitor smaller in volume is completely inserted into the other capacitor larger in volume and kept in symmetric position and then connection polarities of the plates are reversed as shown in figure $$II$$. Identity correct statements ?

A
Capacitance of the new system is $$10 \mathrm { C }$$

B
Capacitance of the new system is $$7 \mathrm { C }$$

C
Charge flown through the battery is $$10 \mathrm { av } $$

D
Charge flown through the battery is $$7 \mathrm { CV }$$