Class 12th
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New answer posted
11 months agoContributor-Level 10
2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge . For positive charge , potential difference V s always positive.
New answer posted
11 months agoContributor-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 agoContributor-Level 10
2.33 Potential rating of the capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, = 3
Dielectric strength = V/m
For safety, the field intensity should not cross 10% of the dielectric strength, hence
Electric field intensity, E = 10 % of = V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 F
Distance between the plates, d is given by d = = m = m
Capacitance is given by the relation
C = , where = permittivity of free space = 8.854
50 =
A = 1.88 = 18.8
Hence, t
New answer posted
11 months agoContributor-Level 10
2.32 Length of the co-axial cylinder, l = 15 cm = 0.15 m
Radius of the outer cylinder, = 1.5 cm = 0.015 m
Radius of the inner cylinder, = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 C
Capacitance of a co-axial cylinder of radii and is given by the relation
C = , where = permittivity of free space = 8.854
C = = 1.21 F
Potential difference of the inner cylinder is given by
V = = = 2.893 V
New answer posted
11 months agoContributor-Level 10
2.30 Radius of the inner sphere, = 12 cm = 0.12 m
Radius of the outer sphere, = 13 cm = 0.13 m
Charges on the inner sphere, q= 2.5 C
Dielectric constant of the liquid, k = 32
Capacitance of the capacitor is given by the relation, C =
= permittivity of free space = 8.854
= 5.55 F
Potential of the inner surface is given by
V = = = 450 V
Radius of the isolate sphere, r = 12 cm = 12 m
Capacitance on the isolated sphere is given by C' = 4 r
= 4 8.854 F
= 1.34 F
The capacitance of the isolated
New answer posted
11 months agoContributor-Level 10
2.29 Let F be the force applied to separate the plates of a parallel plate capacitor and let be the distance.
Hence work done by the force = F
The potential energy increase in the capacitor = uA , 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,
F = uA or F = uA = ( )A
Electric intensity is given by
E =
F = ( )EA= ( )EA
Since Capacitance, C =
F = ( E) = QE
New answer posted
11 months agoContributor-Level 10
2.28 Let F be the force applied to separate the plates of a parallel plate capacitor and let be the distance.
Hence work done by the force = F
The potential energy increase in the capacitor = uA , 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,
F = uA or F = uA = ( )A
Electric intensity is given by
E =
F = ( )EA= ( )EA
Since Capacitance, C =
F = ( E) = QE
New answer posted
11 months agoContributor-Level 10
2.27 Capacitance of the charged capacitor, = 4 F
Supply voltage, = 200 V
Electrostatic energy stored in the capacitor, = = J
= 0.08 J
Capacitance of the uncharged capacitor, = 2 = 2 F
When is connected to , the potential acquired by be
From the conservation of energy, the charge acquired by becomes the charge acquired by and
Hence
or = = = V
Electrostatic energy of the combination of two capacitors is given by
= 
New answer posted
11 months agoContributor-Level 10
2.26 Area of the parallel plate capacitor, A = 90 = 90
Distance between plate, d = 2.5 mm = 2.5 m
Potential difference across plates, V = 400 V
Capacitance of the capacitor is given by, C =
where == 8.854
Electrostatic energy stored in the capacitor is given by the relation
= C = = = 2.55 J
Volume of the given capacitor, V' = A = 90 2.5
= 2.25
Energy stored is given by u = = = 0.113 J/
Also, u = = = =
New answer posted
11 months agoContributor-Level 10
2.25 Let C' be the equivalent capacitance for capacitors and connected in series.
Hence, = + . So C' = 100 pF
Capacitors C' and are in parallel, if the equivalent capacitance be C”, then
C” = C' + = 100 + 100 = 200 pF
Now C” and are connected in series. If the total equivalent capacitance of the circuit be , then = + = + , = pF = F
Let V” be the potential difference across C” and be the potential difference across
Then, V” + 
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