Question

An uniformly charged thin ring has radius $10\text{cm}$ and total charge $12\text{nC}$. An electron is placed on the ring's axis at a distance $25\text{cm}$ from the centre of the ring and is constrained to stay on the axis of ring. The electron is then released from rest. Find the speed of electron when it reaches the center of the ring.

• A. $$1.5\times {10}^4\text{m/s}$$
• B. $$20\times {10}^4\text{m/s}$$
• C. $$18\times {10}^5\text{m/s}$$
• D. $$15.45\times {10}^6\text{m/s}$$

Answer 1

The correct option is D $$15.45\times {10}^6\text{m/s}$$

Given, charge on ring, $q=12\text{nC}$
radius of ring, $r=10\text{cm}$
mass of electron, $m=9.1\times {10}^{-31}\text{kg}$
charge of an electron, $(-e)=-1.6\times {10}^{-19}\text{C}$


As the electron approaches $\text{C}$, its speed (hence, kinetic energy) increases due to force of attraction towards the centre $\text{C}$. This increase in kinetic energy is at the cost of electrostatic potential energy.
$\frac{1}{2}m{v}^2=U_i-U_f...\left(1\right)$
Where,
$v$ is the velocity of electron,
$U_P=-e{V}_P=$electrostatic potential energy of electron at point $\text{P}$,
$U_C=-e{V}_C=$electrostatic potential energy of electron at point $\text{P}$,
Here,
$(-e)$ is the charge of an electron,
Now,
Potential at point $\text{P}$ due to ring,
$V_p=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
Substituting the values, we get
$V_P=\frac{(9\times {10}^9)(12\times {10}^{-9})}{(\sqrt{(10{)}^2+(25{)}^2}\times {10}^{-2})}=401\text{V}$

Potential at point $\text{C}$ due to ring,
$V_C=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

$V_C=\frac{(9\times {10}^9)(12\times {10}^{-9})}{10\times {10}^{-2}}=1080\text{V}$

Substituting the proper values in equation $\left(1\right)$ we have
$\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=(-1.6\times {10}^{-19})(401-1080)$

$\therefore v=15.45\times {10}^6\text{m/s}$

why this question ? Tip: At the initial position (axial position of rings) and final position (centre of ring ) we can find the potential due to ring and hence P.E of electron at both positions. $$P.E\left(U\right)=qV$$