Question

Two conducting sphere of radii 6cm and 12cm each having same charge of 3×10^-8 are kept at surtain distance if each sphere is connected to each other by a conducting wire find direction and amount of charge transfer and potential of each other

Answer 1

Dear Student,

◆ Answer -

Common potential = 3000 V

Charge transfer = 10^-8 C

● Explanation -

Let Q1 & Q2 be charges and V be common potential of each sphere after connecting to each other.

By law of conservation of charge,

Q1 + Q2 = q1 + q2

Q1 + Q2 = 3×10^-8 + 3×10^-8

Q1 + Q2 = 6×10^-8 C ...(1)

Potential will be equalized for both spheres after connection.

V = k.Q1/r1 = k.Q2/r2 ...(2)

Q2 = r2 × Q1 / r1

Q2 = 0.12 × Q1 / 0.06

Q2 = 2Q1 ...(3)

Substitute (3) in (1),

Q1 + 2Q1 = 6×10^-8

3Q1 = 6×10^-8

Q1 = 6×10^-8 / 3

Q1 = 2×10^-8 C

Charge transferred is calculated by -

∆q = q1 - Q1

∆q = 3×10^-8 - 2×10^-8

∆q = 10^-8 C

Final common potential of both spheres is -

V = k.Q1/r1

V = 9×10^9 × 2×10^-8 / 0.06

V = 3×10^3

V = 3000 V

Therefore, common potential is 3000 V and 10^-8 C charge is transferred from small sphere to large sphere...