Electrostatic Potential and Capacitance

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New answer posted•a year ago
S
a year ago

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 

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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&

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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” an

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New answer posted•a year ago
S
a year ago

2.24 Capacitance of the parallel capacitor, C = 2 F

Distance between two plates, d = 0.5 cm = 0.5 ×10-2 m

Capacitance of a parallel plate capacitor is given by the relation,

C = ε0Ad , where ε0=Permittivityoffreespace == 8.854 ×10-12 C2N-1 m-2

A = Cdε0 = 2×0.5×10-28.854×10-12 = 1129 ×106 m = 1129 km

To avoi

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New answer posted•a year ago
S
a year ago

2.23 Requirements:

Capacitance, C = 2 μF

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, C1 = 1 μF

Potential difference, V1 = 400 V

Assumption:

A number of capacitors are connected in series and these series circuits are connected in parallel to each ot

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New answer posted•a year ago
S
a year ago

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ =

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