Electrostatic Potential and Capacitance

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

2.20 For Sphere A: We assume, radius = a, Charge on the sphere = QA , Capacitance = CA , Electric field = EA

For Sphere B: We assume, radius = b, Charge on the sphere = QB , Capacitance = CB , Electric field = EB

EAisgivenby,EA= QA4πε0a2 and EBisgivenby,EB= QB4πε0b2

EAEB = QAQB×b2a2 ……….(

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New answer posted•a year ago
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a year ago

2.19

Let the charge in Proton 1 be q1 = 1.6 ×10-19 C, in Proton 2, q2 = 1.6 ×10-19 C and the charge in the electron, q3 = - 1.6 ×10-19 C

Distance between Proton 1 & Proton 2, d1 = 1.5 Å = 1.5 ×10-10 m

Distance between Proton 1 and Electron, 

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

2.17

Let charge density of the long charged cylinder, with length L and radius r be λ.

Another cylinder of same length surrounds the previous cylinder with radius R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gau

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New answer posted•a year ago
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a year ago

2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

Surface charge density at the inner surface of the shell is given by the relation,

σ1=TotalchargeInnersurfacearea=-q4πr12 …………(1

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New answer posted•a year ago
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a year ago

2.14

Two charges are at point A and B. O is the midpoint of line joining A & B

Magnitude of charge at point A, q1 = 1.5 μC = 1.5 ×10-6C and

at point B, q2 = 2.5 μC=2.5×10-6C

Distance between A & B, d = 30 cm = 0.3 m

Let V1 and E1 be the electric potential and electric

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New answer posted•a year ago
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a year ago

2.13

Length of the side of a cube = b

Charge at each vertices= q

d = diagonal of each face, d = b2+b2 = b √2

l = length of the solid diagonal, l = d2+b2 = b √3

If r is the distance from the centre of the cube to its corner, then r = l/2 = √32b

The electric potential (V) at the centre of t

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New answer posted•a year ago
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a year ago

2.12

Charge located at the origin O, q=8 mC= 8 ×10-3 C

Magnitude of a small charge, which is taken from point P to Q, q1 = -2 ×10-9 C

Pont P is at a distance, d1 = 3 cm = 0.03 m from origin, along Z axis

Point Q is at a distance, d2 = 4 cm = 0.04 m from origin, along y

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New answer posted•a year ago
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a year ago

2.11 Capacitance of the first capacitor, C = 600 pF = 600 ×10-12 F

Potential difference, V = 200V

Electrostatic energy of the first capacitor, E1 = 12 C V2 = 12× 600 ×10-12×2002 J

= 1.2 ×10-5 J

After the supply is disconnected and the first capacitor is connected t

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