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New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2 . For positive charge q1 , potential difference V s always positive.

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.

Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases

Concentric spheres centered at the origin are equipotential surfaces.

A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

2.33 Potential rating of the capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, ?r = 3

Dielectric strength = 107 V/m

For safety, the field intensity should not cross 10% of the dielectric strength, hence

Electric field intensity, E = 10 % of 107 = 106 V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 *10-12 F

Distance between the plates, d is given by d = VE = 1000106 m = 10-3 m

Capacitance is given by the relation

C = ε0?rAd , where ε0 = permittivity of free space = 8.854 *10-12 C2N-1 m-2

50 *10-12 = 8.854*10-12*3*A10-3

A = 1.88 *10-3 m2 = 18.8 cm2

Hence, t

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New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

2.32  Length of the co-axial cylinder, l = 15 cm = 0.15 m

Radius of the outer cylinder, r1 = 1.5 cm = 0.015 m

Radius of the inner cylinder, r2 = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5 μC=3.5*10-6 C

Capacitance of a co-axial cylinder of radii r1 and r2 is given by the relation

C = 2πε0lloge?r1r2 , where ε0 = permittivity of free space = 8.854 *10-12 C2N-1 m-2

C = 2π*8.854*10-12*0.15loge?0.0150.014 = 1.21 *10-10 F

Potential difference of the inner cylinder is given by

V = qC = 3.5*10-61.21*10-10 = 2.893 *104 V

New answer posted

11 months ago

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A
alok kumar singh

Contributor-Level 10

2.30 Radius of the inner sphere, r2 = 12 cm = 0.12 m

Radius of the outer sphere, r1 = 13 cm = 0.13 m

Charges on the inner sphere, q= 2.5 μC=2.5*10-6 C

Dielectric constant of the liquid, k = 32

Capacitance of the capacitor is given by the relation, C = 4πε0kr1r2(r1-r2)

ε0 = permittivity of free space = 8.854 *10-12 C2N-1 m-2

C= 4π*8.854*10-12*32*0.13*0.12(0.13-0.12) = 5.55 *10-9 F

Potential of the inner surface is given by

V = qC = 2.5*10-65.55*10-9 = 450 V

Radius of the isolate sphere, r = 12 cm = 12 *10-2 m

Capacitance on the isolated sphere is given by C' = 4 πε0 r

= 4 *π* 8.854 *10-12*12*10-2 F

= 1.34 *10-11 F

The capacitance of the isolated

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New answer posted

11 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

2.29 Let F be the force applied to separate the plates of a parallel plate capacitor and let Δx be the distance.

Hence work done by the force = F Δx

The potential energy increase in the capacitor = uA Δx , where u = Energy density, A = area of each plate.

If d = distance between the plates and V = potential difference across the plates, the

Work done = increase in the potential energy.

Therefore,

Δx = uA Δx or F = uA = ( 12ε0E2 )A

Electric intensity is given by

E = Vd

F = ( 12ε0E )EA= ( 12ε0Vd )EA

Since Capacitance, C = ε0Ad

F = ( 12CV E) = 12 QE

New answer posted

11 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

2.28 Let F be the force applied to separate the plates of a parallel plate capacitor and let Δx be the distance.

Hence work done by the force = F Δx

The potential energy increase in the capacitor = uA Δx , where u = Energy density, A = area of each plate.

If d = distance between the plates and V = potential difference across the plates, the

Work done = increase in the potential energy.

Therefore,

Δx = uA Δx or F = uA = ( 12ε0E2 )A

Electric intensity is given by

E = Vd

F = ( 12ε0E )EA= ( 12ε0Vd )EA

Since Capacitance, C = ε0Ad

F = ( 12CV E) = 12 QE

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2.27 Capacitance of the charged capacitor, C1 = 4 μF=4*10-6 F

Supply voltage, V1 = 200 V

Electrostatic energy stored in the C1 capacitor, E1 = 12C1V12 = 12*4*10-6*2002 J

= 0.08 J

Capacitance of the uncharged capacitor, C2 = 2 μF = 2 *10-6 F

When C2 is connected to C1 , the potential acquired by C2 be V2

From the conservation of energy, the charge acquired by C1 becomes the charge acquired by C1 and C2.

Hence V2*C1+C2=C1V1

or V2 = C1V1C1+C2 = 4*10-6*200(4*10-6+2*10-6) = 4003 V

Electrostatic energy of the combination of two capacitors is given by

E2 = 

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New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

2.26 Area of the parallel plate capacitor, A = 90 cm2 = 90 *10-4 m2

Distance between plate, d = 2.5 mm = 2.5 *10-3 m

Potential difference across plates, V = 400 V

Capacitance of the capacitor is given by, C = ε0Ad,

where ε0=Permittivityoffreespace == 8.854 *10-12 C2N-1 m-2

Electrostatic energy stored in the capacitor is given by the relation

E1 = 12 C V2 = 12 *ε0Ad*V2 = 12 *8.854*10-12*90*10-42.5*10-3*4002 = 2.55 *10-6 J

Volume of the given capacitor, V' = A *d = 90 *10-4* 2.5 *10-3 m3

= 2.25 *10-5 m3

Energy stored is given by u = E1V' = 2.55*10-62.25*10-5 = 0.113 J/ m3

Also, u = E1V' = 12CV2Ad = 12*ε0Ad*V2Ad =

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New answer posted

11 months ago

0 Follower 68 Views

A
alok kumar singh

Contributor-Level 10

2.25 Let C' be the equivalent capacitance for capacitors C2 and C3 connected in series.

Hence, 1C' = 1200 + 1200 . So C' = 100 pF

Capacitors C' and C1 are in parallel, if the equivalent capacitance be C”, then

C” = C' + C1 = 100 + 100 = 200 pF

Now C” and C4 are connected in series. If the total equivalent capacitance of the circuit be CTotal , then 1CTotal = 1C" + 1C4 = 1200 + 1100 , CTotal = 2003 pF = 2003*10-12 F

Let V” be the potential difference across C” and V4 be the potential difference across C4

Then, V” + 

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