Question

An electron is projected from the right plate of set I directly towards its left plate. It just comes to rest at the plate. The speed with which it was projected is about:
(Take \( \frac{e}{m} = 1.76 \times 10^{11} \, C/kg} \))

• A. \( 1.3 ×10^5m/s\)
• B. \(2.6 ×10^6m/s\)
• C. \(6.5 ×10^5m/s\)
• D. \(5.2 ×10^7m/s\)

Hint:

When projecting an electron or any charged particle in an electric field, remember that kinetic energy is converted into electrical potential energy, and vice versa. The greater the potential difference, the faster the initial speed required to just reach the opposite plate.

Answer 1

The Correct Option is B

Step 1: The work done on the electron by the electric field between the plates is equal to the change in its kinetic energy.
The work done by the electric field is given by:
\[ W = eV \] where \( e \) is the charge of the electron and \( V \) is the potential difference between the plates.
The initial kinetic energy of the electron is \( \frac{1}{2} m v^2 \), and the final kinetic energy is zero, since it just comes to rest.
Using the work-energy theorem, the work done by the electric field is equal to the change in kinetic energy:
\[ eV = \frac{1}{2} m v^2 \] Now, from the previous part (29), the potential difference between the plates of set I is \( 20 \, V} \). Step 2: We can calculate the velocity of the electron using the relation:
\[ v = \sqrt{\frac{2 eV}{m}} \] Substitute the values:
\[ v = \sqrt{\frac{2 \times 1.76 \times 10^{-19} \times 20}{9.11 \times 10^{-31}}} \] \[ v \approx 2.6 \times 10^6 \, m/s} \]