Question

A proton travels through a distance of $5\text{ m}$ in the direction of uniform electric field of intensity $4\text{ NC}^{-1}$. The work done on the proton by the electric field is

• A. 10 eV
• B. 30 eV
• C. 20 eV
• D. 16 eV
• E. 32 eV

Hint:

The work done in eV is numerically equal to the product of the field intensity and distance if the charge is $1e$. Here, $4 \times 5 = 20 \text{ eV}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Work done by a force is given by $W = \vec{F} \cdot \vec{d}$. For a charge $q$ in an electric field $\vec{E}$, the force is $\vec{F} = q\vec{E}$.

Step 2: Detailed Explanation:
1. Magnitude of force on the proton: $F = qE$.
2. Since the proton moves in the direction of the field, the angle $\theta = 0^\circ$.
Work done $W = Fd \cos 0^\circ = qEd$.
3. Substitute values:
Charge of proton $q = +e$ (elementary charge).
Electric field $E = 4\text{ NC}^{-1}$.
Distance $d = 5\text{ m}$.
\[ W = e \times 4 \times 5 = 20e \text{ Joules} \]
4. Convert Joules to electron-volts (eV):
$1\text{ eV} = e \text{ Joules}$.
Therefore, $20e \text{ Joules} = 20\text{ eV}$.

Step 3: Final Answer:
The work done is 20 eV.