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 \( d=0.25 \text{ m} \):
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 \( d=0.25 \text{ m} \):
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For perpendicular charged rods, calculate the field magnitude of each rod individually and use vector addition to find the total field.
Updated On: Jun 9, 2026
- \( 450 \text{ Vm}^{-1} \)
- \( 568 \text{ Vm}^{-1} \)
- \( 406 \text{ Vm}^{-1} \)
- \( 203 \text{ Vm}^{-1} \)
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The Correct Option is C
Solution and Explanation
Concept:
The electric field \( E \) at a distance \( d \) from one end of a uniformly charged rod of length \( L \) is given by \( E = \frac{kQ}{d(d+L)} \), where \( k = \frac{1}{4\pi\epsilon_0} \).
Step 1: Calculate the field due to one rod.
\( Q = 10^{-8} \text{ C} \), \( L = 1 \text{ m} \), \( d = 0.25 \text{ m} \). $$ E_{rod} = \frac{9 \times 10^9 \times 10^{-8}}{0.25(0.25 + 1)} $$ $$ E_{rod} = \frac{90}{0.25 \times 1.25} = \frac{90}{0.3125} = 288 \text{ Vm}^{-1} $$
Step 2: Consider both rods.
There are two perpendicular rods producing fields of magnitude \( 288 \text{ Vm}^{-1} \) in orthogonal directions. The resultant electric field is \( E_{net} = \sqrt{E_1^2 + E_2^2} \): $$ E_{net} = \sqrt{288^2 + 288^2} = 288 \sqrt{2} \approx 288 \times 1.414 \approx 407.2 \text{ Vm}^{-1} $$
Step 3: Final determination.
The value matches approximately 406 \( \text{Vm}^{-1} \). $$\boxed{406 \text{ Vm}^{-1}}$$
Step 1: Calculate the field due to one rod.
\( Q = 10^{-8} \text{ C} \), \( L = 1 \text{ m} \), \( d = 0.25 \text{ m} \). $$ E_{rod} = \frac{9 \times 10^9 \times 10^{-8}}{0.25(0.25 + 1)} $$ $$ E_{rod} = \frac{90}{0.25 \times 1.25} = \frac{90}{0.3125} = 288 \text{ Vm}^{-1} $$
Step 2: Consider both rods.
There are two perpendicular rods producing fields of magnitude \( 288 \text{ Vm}^{-1} \) in orthogonal directions. The resultant electric field is \( E_{net} = \sqrt{E_1^2 + E_2^2} \): $$ E_{net} = \sqrt{288^2 + 288^2} = 288 \sqrt{2} \approx 288 \times 1.414 \approx 407.2 \text{ Vm}^{-1} $$
Step 3: Final determination.
The value matches approximately 406 \( \text{Vm}^{-1} \). $$\boxed{406 \text{ Vm}^{-1}}$$
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