Electrostatic Potential and Capacitance

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

12 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

As we know that, $σ$ =charge/Area

Q₁= $σ$ (4 $π r$ ²)

Q₂= $σ$ (4 $π (2 r)$ ²)= 4Q₁

when they come in contact with each other potential being same and charge will be conserved.

So net charge is 5Q₁= 5 ( $σ$ (4 $π r$ ²)

Also V₁=V₂

KQ₁/R=KQ₂/2R so Q₁=Q₂/

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3)e] and two down quarks [charges (-l/3)e]. Assume that they have a triangle configuration with side length of the order of 10^-15 Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.

(b) Repeat above exercise for a proton which is made of two up and one down quark.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

Potential energy of system of charges be algebraic sum

U = $\frac{1}{4 π ? ?} (\frac{q q}{r} - \frac{q q}{r} - \frac{q q}{r})$

9 $\times$ 10⁹/10^-15 (1.6 $\times 10$ ^-19)² (1/3)²- (2/3) (1/3)- (2/3) (1/3)

-7.68 $\times$ 10^-14J

-7.68 $\times$ 10^{-14 $\times 1.6 \times 10$}^-19= 0.48Mev

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A capacitor is made of two circular plates of radius R each, separated by a , distance d << R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r << R and thickness t << r is placed at the centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $?$ _{0 $\frac{v}{d} ?$}r², using this relation in 1 and 2

F=- $?$ _{0 $\frac{v}{d} ?$}r²V/d…… (3)

Also to balance something F=mg……. (

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε= αU where α = 2V^-1. A similar capacitor with no dielectric is charged to U₀= 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $\in$ _{0 $\frac{v}{d} π$}r², using this relation in 1 and 2

F=- $\in$ _{0 $\frac{v}{d} π$}r²V/d…… (3)

Also to balance something F=mg……. (

Upvote

New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

Two point charges of magnitude +q and -q are placed at (-d/2, 0, 0) and (d/2, 2, 0), respectively. Find the equation of the equipotential surface where the potential is zero

18 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\frac{1}{4 π ? ?} \frac{q}{(\sqrt[]{x^{2} + {(d / 2)}^{2} + h^{2}}}$ - $\frac{1}{4 π ? ?} \frac{q}{(\sqrt[]{x^{2} - {(d / 2)}^{2} + h^{2}}}$

If net potential is zero then,

$\frac{1}{(\sqrt[]{x^{2} + {(d / 2)}^{2} + h^{2}}}$ = $\frac{1}{(\sqrt[]{x^{2} + {(d / 2)}^{2} + h^{2}}}$

x^{2} + {(d / 2)}^{2} + h^{2}

x^{2} - {(d / 2)}^{2} + h^{2}

2dx= 0 , x=0

X=0 , y-z plane.

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New answer posted•a year ago

Electrostatic Potential and Capacitance physics ncert exemplar solutions class 12th chapter two

Find the equation of the equipotential for an infinite cylinder of radius r₀ carrying charge of linear density A.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Electrostatic Potential and Capacitance Class 12th

2.36 Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm–1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?

(Hint: The earth has an electric field of about 100 Vm–1 at its surface in the downward direction, corresponding to a surface charge density = –10–9 C m–2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

27 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

2.36 We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

Yes, the person will get an electric shock if he touches the metal slab next morning. The steady discharging current

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New answer posted•a year ago

Electrostatic Potential and Capacitance Class 12th

2.35 A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.

79 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge $q_{1}$ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge $q_{2}$ . For positive charge $q_{1}$ , potential difference V