A uniformly charged thin ring has radius 15.0 cm and total charge 23.5 nC. An electron is placed on the ring's axis a distance 30.5 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. Describe the subsequent motion of the electron. Find the speed of the electron when it reaches the center of the ring.

Explanation:

The motion of the electron will be periodic oscillation  , ie it will go to and fro about the center of the ring. It will not be SHM as the distance at which electron has been placed is far off from the center.

Potential at the center = 9 x 10⁹ x 23.5 x 10⁻⁹ / .15

= 1410 V

Potential at distance of 30.5 cm

= 9 x 10⁹ x 23.5 x 10⁻⁹ / √ (15² +30.5²)

=9 x 10⁹ x 23.5 x 10⁻⁹ / .40

= 528.75 V

according to conservation of energy

1/2 mv² = q ( v₂ - v₁ )

1/2 x 9.1 x 10⁻³¹ x v² = (1410 - 528.75 ) x 1.6 x 10⁻¹⁹

v² = 309.9 x 10¹²

v = 17.6 x 10⁶ m /s

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energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j