Question

Two similar rods of length \( l=1 \text{ m} \) carrying equal charges \( (q) = 10^{-8} \text{ C} \) are placed as shown in figure. The electric field at point 'O' approximately is, if \( d=0.25 \text{ m} \):

• A. \( 450 \text{ Vm}^{-1} \)
• B. \( 568 \text{ Vm}^{-1} \)
• C. \( 406 \text{ Vm}^{-1} \)
• D. \( 203 \text{ Vm}^{-1} \)

Hint:

For perpendicular charged rods, calculate the field magnitude of each rod individually and use vector addition to find the total field.

Answer 1

The Correct Option is C

Step 1: Field of one charged rod.
A uniformly charged rod of length $L$ produces, at an axial point a distance $d$ from its near end, a field $E = \dfrac{kQ}{d(d+L)}$, with $k = 9\times10^{9}$.
Step 2: List the data.
$Q = 10^{-8}\,\text{C}$, $L = 1\,\text{m}$, $d = 0.25\,\text{m}$.
Step 3: Field due to one rod.
\[ E_{\text{rod}} = \frac{9\times10^{9}\times10^{-8}}{0.25\,(0.25+1)} = \frac{90}{0.25\times1.25} = \frac{90}{0.3125} \approx 288\,\text{V m}^{-1}. \]
Step 4: Two rods, perpendicular fields.
By the geometry shown, the two identical rods send fields of about $288\,\text{V m}^{-1}$ each at point $O$, directed at right angles to one another.
Step 5: Combine at right angles.
Perpendicular fields add as a vector sum: \[ E_{\text{net}} = \sqrt{288^2 + 288^2} = 288\sqrt{2}. \]
Step 6: Evaluate.
$288 \times 1.414 \approx 407\,\text{V m}^{-1}$, which matches the option $406\,\text{V m}^{-1}$.
\[ \boxed{406\ \text{V m}^{-1}} \]