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 on a charged particle by a uniform electric field is the product of the force and the displacement in the direction of the force. We then convert this work from Joules to electron-volts (eV).
Step 2: Key Formula or Approach:
Force on a charge: \(F = qE\).
Work done: \(W = F \times d = qEd\).
To convert from Joules to eV, divide by the elementary charge \(e\) (since \(1 \text{ eV} = e \text{ Joules}\)).
Step 3: Detailed Explanation:
Given values:
Charge of proton \(q = +e = 1.6 \times 10^{-19} \text{ C}\)
Electric field \(E = 4 \text{ N/C}\)
Distance \(d = 5 \text{ m}\)
Calculate work in Joules:
\[ W = (e)(4)(5) = 20e \text{ Joules} \] Convert Joules to electron-volts (eV). We divide the energy in Joules by the elementary charge \(e\):
\[ W_{\text{eV}} = \frac{20e \text{ Joules}}{e \text{ Joules/eV}} = 20 \text{ eV} \] Step 4: Final Answer:
The work done is 20 eV.