Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 74

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Weekly Quiz Competition

Question 1

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Match the following two columns:

	Column I		Column II
A.	Electrical resistance	$$1.$$	$$[ML^{3}T^{-3}A^{-2}]$$
B.	Electrical potential	$$2.$$	$$[ML^{2}T^{-3}A^{-2}]$$
C.	Specific resistance	$$3.$$	$$[ML^{2}T^{-3}A^{-1}]$$
D.	Specific conductance	$$4.$$	None of these

$$A - 2, B - 3, C - 1, D - 4$$

$$A - 2, B - 4, C - 3, D - 1$$

$$A - 1, B - 2, C - 3, D - 3$$

$$A - 1, B - 3, C - 2, D - 4$$

Solution

The dimensions of electrical resistance
$$R = \dfrac {V}{I} = \dfrac {\left (\dfrac {W}{q}\right )}{I} = \dfrac {\dfrac {W}{It}}{(I)}$$
$$= \dfrac {W}{I^{2}t} = [ML^{2} T^{-2}T^{-1}A^{-2}]$$
$$= [ML^{2}T^{-3}A^{-2}]$$
Then, $$(A) \rightarrow (2)$$
The dimensions of electrical potential
$$V = \dfrac {W}{q} = \dfrac {W}{It} = [ML^{2}T^{-2}A^{-1}T^{-1}]$$
$$= [ML^{2}T^{-3}A^{-1}]$$
Then, $$(B)\rightarrow (3)$$
The dimensions of specific resistance
$$\rho = R \dfrac {l}{A} = [ML^{2}T^{-3}A^{-2}][L]$$
$$= [ML^{3}T^{-3}A^{-2}]$$
Thus, $$(C) \rightarrow (2)$$ and the dimensions of specific conductance
$$\sigma = \dfrac {1}{\rho} = \dfrac {1}{[ML^{3}T^{-3}A^{-2}]}$$
$$= [M^{-1}L^{-3}T^{3}A^{2}]$$
$$=$$ not give in column
Thus, $$(D) \rightarrow (4)$$.

Question 2
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A very thin metal sheet is inserted halfway between the parallel plates of an air-gap capacitor. The sheet is thin compared to the distance between the plates, and it does not touch either plate when fully inserted. The system had capacitance, $$C$$, before the plate is inserted.
What is the equivalent capacitance of the system after the sheet is fully inserted?

A
$$\cfrac{1}{4}C$$

B
$$\cfrac{1}{2}C$$

C
$$C$$

D
$$2C$$

E
$$4C$$

Solution

Initially (before metal sheet inserted) the capacitance of a parallel plate capacitor is $$C=\dfrac{A\epsilon_0}{d}$$ where A be the area of plates and d be the separation between parallel plates.
When a metal sheet inserted fully halfway between the parallel plates, the capacitance will be divided into two capacitors $$C_1, C_2$$ and they are in series.
Thus, $$C_1=\dfrac{A\epsilon_0}{(d/2)}=2C$$ and $$C_2=\dfrac{A\epsilon_0}{(d/2)}=2C$$
The equivalent capacitance , $$C_{12}=\dfrac{C_1C_2}{C_1+C_2}=\dfrac{2C\times 2C}{2C+2C}=C$$
Question 3
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The plates of a parallel plate capacitor are charged upto $$100 V$$.A $$2 mm$$ thick insulator shunt is inserted between the plates . Then to maintain the same potential difference ,the distance between the same potential difference , the distance between the capacitor plates is increased by $$ 1.6 mm$$.The dielectric constant of the insulator is

A
$$5$$

B
$$8$$

C
$$4$$

D
$$6$$

Solution

Here,$$V=100 V$$ and $$ t=2 mm$$
As $$ V=\dfrac{q}{c}$$ and $$ V$$ and $$ q $$ remain unchanged ,
therefore C must be constant
ie, $$ C_i=C_f$$
$$\dfrac{\varepsilon_0A}{d}=\dfrac{\varepsilon_0\,A}{d+1.6-t+\dfrac{t}{k}}$$
$$\therefore 1.6-t+\dfrac{t}{k}=0$$
As $$t=2 mm$$,therefore
$$1.6-2+\dfrac{2}{k}=0$$
or $$\dfrac{2}{k}=0.4$$
$$\therefore k=\dfrac{2}{0.4}=5$$
Question 4
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A uniform electric field of magnitude $$290 V/m$$ is directed in the positive $$x$$ direction. A $$+13.0 \mu C$$ charge moves from the origin to the point $$(x, y) = (20.0 cm, 50.0 cm).$$

What is the change in the potential energy of the charge field system?

A
$$-754J$$

B
$$-754mJ$$

C
$$-754kJ$$

D
$$-754\mu J$$

Solution

Given electric field, $$\overrightarrow{E}=290\, V/m$$ directed along $$+x$$ direction.
Charge $$=+13\mu C$$ moves from origin to point $$(x,y)=(20\, cm , 50\, cm)$$.
We have to find the charge in potential energy of the charge field system.
We know, the change in potential energy $$=$$ charge $$\times $$ change in potential
i.e, $$\Delta U=q\Delta V$$
But, $$\Delta V=-Ed$$
$$\Delta U=-qEd$$
$$=-13\times 10^{-6}\times 290\times 20\times 10^{-2}$$
$$=0.000754\, J$$
$$=-754\times 10^{-6}\, J$$
Here $$d=20\, cm$$, since electric field is along positive $$x-$$axis.
Question 5
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The area of the plates of a parallel plate capacitor is $$A$$ and the gap between them is $$d$$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $$'y'$$from one plate as : $$K\ =\ \lambda \sec \left(\frac{\pi y}{2d}\right)$$ , where $$\lambda$$ is a dimensionless constant. The capacitance of this capacitor is

A
$$\large{\frac{\pi \epsilon_0 \lambda A}{2d}}$$

B
$$\large{\frac{\pi \epsilon_0 \lambda A}{d}}$$

C
$$\large{\frac{2\pi \epsilon_0 \lambda A}{d}}$$

D
None

Solution

$$E_y = \cfrac{\sigma}{ky\epsilon_0}$$

$$dV = \cfrac{\sigma}{k\epsilon_0} dy$$

$$dV = \cfrac{\sigma}{\lambda \epsilon_0} \cos {(\cfrac{\pi y}{2d})}dy$$

Integrating to dy and other side;

$$\int^{V}_{0} dV = \int^{d}_{0} \cfrac{\sigma}{\lambda \epsilon_0} \cos{(\cfrac{\pi y}{2d})} dy$$

$$V = \cfrac{\sigma}{\lambda \epsilon_0} . \cfrac{2d}{\pi}$$

$$C = \cfrac{Q}{\sigma}(\cfrac{\lambda \epsilon_0 \pi}{2d})$$

$$C = \cfrac{A \lambda \epsilon_0 \pi}{2d}$$
Question 6
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Five conducting plates are placed parallel to each other Separation between them is $$d$$ and area of each plate is $$A$$. Plate number $$1,2$$ and $$3$$ are connected with each other and at the same time through a cell of emf $$E$$. The charge on plate number $$1$$ is

A
$$E{ \varepsilon }_{ 0 }A/d$$

B
$$E{ \varepsilon }_{ 0 }A/2d$$

C
$$2E{ \varepsilon }_{ 0 }A/d$$

D
zero

Solution

Since plates $$1,2,3$$ are connected in parallel.
$$C_{eq}=C_1+C_2+C_3=\cfrac{3\epsilon_oA}{d}$$
Toatl charges $$Q=CE=\cfrac{3\epsilon_oAE}{d}$$
so, charge through each plates$$=\cfrac{Q}{3}=\cfrac{\epsilon_oAE}{d}$$
Question 7
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A capacitor of capacitance $${ C }_{ 1 }=1\mu F$$ can with stand maximum voltage $${ V }_{ 1 }=6kV$$ (kilo-volt) and another capacitor of capacitance $${ C }_{ 2 }=3\mu F$$ can withstand maximum voltage $${ V }_{ 2 }=4kV$$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of:

A
$$4kV$$

B
$$6kV$$

C
$$8kV$$

D
$$10kV$$
Question 8
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In a quark model of elemetary particles, a neutron is made of one up quark of charge $$\cfrac{2}{3}e$$ and two down quark of charges $$\left( -\cfrac { 1 }{ 3 } e \right) $$. If they have a triangle configuration with side length of the order of $${ 10 }^{ -15 }m$$. The electrostatic potential energy of neutron in $$MeV$$ is

A
$$7.68$$

B
$$-5.21$$

C
$$-.048$$

D
$$9.34$$

Solution

Fig shows the quark model of a neutral, where $$r=10^{-15}m$$.
Potential energy of neutron
$$V=\dfrac{1}{4\pi \epsilon_0 r}[q_eq_d+q_uq_d+q_uq_d]$$

$$=\dfrac{9\times 10^9}{10^{-15}}\left[\left(-\dfrac{e}{3}\right)\left(-\dfrac{e}{3}\right)+\left(\dfrac{2e}{3}\right)\left(-\dfrac{e}{3}\right)+\left(\dfrac{2e}{3}\right)\left(-\dfrac{e}{3}\right)\right]$$

$$V=\dfrac{9\times 10^9}{10^{-15}}\left(\dfrac{1}{9}-\dfrac{4}{9}\right)(1.6\times 10^{-19})^2=-7.68\times 10^{-14}J$$

$$V=\dfrac{-7.68\times 10^{-14}}{1.6\times 10^{-19}}eV=-4.8\times 10^5eV=-0.48MeV$$
Question 9
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A solid conducting sphere of radius $${R_1}$$ is surrounded by another concentric hollow conducting sphere of radius $${R_2}$$. the capacitance of this assembly is proportional to:-

A
$$\dfrac{{{R_2} - {R_1}}}{{{R_1}{R_2}}}$$

B
$$\dfrac{{{R_2} + {R_1}}}{{{R_1}{R_2}}}$$

C
$$\dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$$

D
$$\dfrac{{{R_1}{R_2}}}{{{R_2} - {R_1}}}$$

Solution
Question 10
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The gravitational field strength $$\vec {E}$$ and gravitational potential $$V$$ are related as $$\vec {E} = -\left (\dfrac {\partial V}{\partial x} \hat {i} + \dfrac {\partial V}{\partial y} \hat {j} + \dfrac {\partial V}{\partial z}\hat {k}\right )$$
In the figure, transversal lines represent equipotential surfaces. A particle of mass $$m$$ is released from rest at the origin. The gravitational unit of potential, $$1\overline {V} = 1\ cm^{2}/ s^{2}$$.
Speed of the particle $$(v) (y$$ is in cm $$v$$ is in $$cm/s$$) as function of its y-co-ordinate is

A
$$v = 2\sqrt {y}$$

B
$$v = \sqrt {2}. y$$

C
$$v = 2y$$

D
$$v = 2y + 4y^{2} + 2y^{2}$$