A ring of radius 6cm is given a charge 10 micro coulomb. How much work will be done in transporting a charge of 6 micro coulomb from its centre to a point 8cm along it's axis .?

Question

Answer

Given:

Radius of the ring, r = 6 cm

Charge on the ring, Q = 10 micro coulomb

Charge to be transported, q = 6 micro coulomb

Distance from the center of the ring to the final point, d = 8 cm

Calculations:

Electric potential at the center of the ring can be calculated as:

V = kQ/r

V = (9 × 10^9 Nm^2/C^2) × (10 × 10^-6 C) / (6 × 10^-2 m)

V = 1.5 × 10^6 V

The work done in moving a charge q from a point P with potential V1 to a point Q with potential V2 is given by:

W = q(V2 - V1)

Electric potential at a point P at a distance x from the center of the ring can be calculated as:

V = kQ/sqrt(r^2 + x^2)

Electric potential at the final point, V2 can be calculated as:

V2 = kQ/sqrt(r^2 + d^2)

V2 = (9 × 10^9 Nm^2/C^2) × (10 × 10^-6 C) / sqrt((6 × 10^-2 m)^2 + (8 × 10^-2 m)^2)

V2 = 8.29 × 10^5 V

Electric potential at the center of the ring, V1 = 1.5 × 10^6 V

Work done in moving the charge from the center of the ring to the final point:

W = q(V2 - V1)

W = (6 × 10^-6 C) × (8.29 × 10^5 V - 1.5 × 10^6 V)

W = -7.02 × 10^-1 J

Answer:

The work done in transporting a charge of 6 micro coulomb from the center of the ring to a point 8cm along its axis is -0.702 J. The negative sign indicates that the work is done against the electric field of the ring.

Answer

5.4 j

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