Question

3) A ring of radius 15 cm is injected with an amount of charge of 3.5 × 10−12C, uniformly distributed along the ring. The ring rotates with an angular velocity of 17 rad/s. Find the power radiated by the system.

Answer 1

Answer:

Explanation:

Given that

A ring of radius 15cm

r=0.15m

A charge of Q=3.5×10^-12C

Angular velocity is given as

w=17rad/s

Power is given as

P=IV, Electrical power

And the rate of charge I is given as

I=Q/t

Also, the potential difference V, for a point charge is given as

V=kQ/r

Therefore, power can be rewritten as

P=IV

P=(Q/t) × (kQ/r)

P=kQ²/tr

Note, t is the period and is given as

w=2πf

Then, f=1/t

w=2π/t,

So, 1/t=w/2π

P=kQ²/tr

P=(kQ²/r)(1/t) = (kQ²/r)(w/2π)

P=(kQ²/r)(w/2π)

K is a constant and it value is 9×10^9Nm²/C²

P=(9×10^9×3.5×10^-12×3.5×10^-12)(17/2π)

P=1.1025×10^-13×17/2π

P=2.98×10^-13 Watts

Answer 2

Given Information:

Radius = r = 15 cm = 0.15 m

Charge = q = 3.5x10⁻¹² C

Angular velocity = ω = 17 rad/s

Required Information:

Power radiated = P = ?

Answer:

Power radiated = 1.987x10⁻¹² W

Explanation:

We know that power is given by

P =  VI

Where V is the potential due to charge and is given by

V = kq/r

Where k is the coulomb constant 8.98x10⁹ N.m²/C²

V = (8.98x10⁹*3.5x10⁻¹²)/0.15

V = 0.2095 V

we know that the rate of charge flow is current

I = q/t

where t is found by the relation

ω = 2π/t

t = 2π/ω

t = 2π/17

t = 0.369 s

so current is

I = q/t = 3.5x10⁻¹²/0.369

I = 9.485x10⁻¹² A

Therefore, the amount of power radiated is

P =VI

P = 0.2095*9.485x10⁻¹²

P = 1.987x10⁻¹² W