Question

A ring of radius 0.1m is made out of a thin metallic wire of area of cross- section 10-6m2. The ring has a uniform charge of coulomb. Find the change in the radius of the ring when a charge of 10-8 coulomb is placed at the centre of the ring. Youngs modulus of the metal is 2x1011 n/m2 .

Answer 1

Answer: 2.25mm

Explanation: let us consider that an circular arc element of charge dq on ring so, when we place an charge Q at centre of the ring then due to electrostatic repulsion b/w charges an extra tension is acting in ring as shown in figure.

so, Kq(dq)/R² = 2dTsinΘ

for small value of Θ , sinΘ ≈ Θ

and also dq = Q/2πR

so, dT = KqQ/2πR² .............(1)

now due to ring is metal so, stress developed in ring.

e.g., stress = Y × strain

dT/A = Y × dR/R

from equation (1),

dR = KqQ/2πRAY

now putting values of all given terms .

e.g., K = 9 × 10⁹ Nm²/C², q = π coulombs Q = 10⁻⁸ coulombs

R = 0.1 m, A = 10⁻⁶ m² and Y = 2 × 10¹¹ N/m²

so, dR = 0.00225 m = 2.25 mm

Answer 2

A ring is having uniformally distributed charge on it. Let the magnitude of this charge be q1.

And now placing a charge q2 at its center, due to the same polarity, because of repulsion the ring will have the tendency to expand and there will be increase in tension of the ring.

Tension can be given as the sum of scalar radial force.

i.e T = F(radial) / 2π

or, T = kq1q2/2πr²

Substituting the values,

T = 9*10^9*10^-8 * π / 2π(0.1)²

or, T = 4500 N

Now, The intial radius is r and after expansion it increases by Δr.

Then, Strain in ring = Δr / r

And, Young's modulus is given as:

Y = stress / strain

or, Y = ( T / S) / (Δr / r)

where, S = cross sectional area

Increase in radius of ring = Δr = Tr / YS

or, Δr = (4500*0.1) / (2*10^11*10^-6)

or, Δr = 2.25 * 10^-3 m

or, Δr = 2.25 mm

Hence, Increase in radius is 2.25 mm.