Question

In hydrogen atom an electron revolves around a proton (in nucleus) at a distance 'r' m. The intensity of electric field due to the proton at distance 'r' is $5 \times 10^{11}\ \text{NC}^{-1}$, the magnitude of force between the electron and proton is [charge on electron = $1.6 \times 10^{-19}\ \text{C}$]

• A. $4 \times 10^8\ \text{N}$
• B. $8 \times 10^8\ \text{N}$
• C. $4 \times 10^{-8}\ \text{N}$
• D. $8 \times 10^{-8}\ \text{N}$

Hint:

Don't waste time using Coulomb's law ($\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$) to solve for the radius $r$. The electric field term $E$ already encapsulates the constant and the distance factor ($\frac{1}{4\pi\varepsilon_0}\frac{q_{\text{proton}}}{r^2}$), so a single direct multiplication ($F = qE$) is all that is required.

Answer 1

The Correct Option is D

Step 1: Understand the situation.
The proton sets up an electric field at the electron's location, and that field exerts a force on the electron. We are given the field strength and the electron charge.
Step 2: Recall the meaning of electric field.
Electric field is force per unit charge, so the force on a charge placed in a field is $F = qE$.
Step 3: List the data.
$E = 5 \times 10^{11}\,\text{N C}^{-1}$ and $q = 1.6 \times 10^{-19}\,\text{C}$.
Step 4: Substitute.
$F = (1.6 \times 10^{-19})(5 \times 10^{11})$.
Step 5: Multiply the numbers.
$1.6 \times 5 = 8.0$.
Step 6: Combine the powers of ten.
$10^{-19} \times 10^{11} = 10^{-8}$, giving $F = 8 \times 10^{-8}\,\text{N}$, which is option (4).
\[ \boxed{F = 8 \times 10^{-8}\ \text{N}} \]