A 10cm long rod carries a charge of +50μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10cm from both the ends of the rod.
A
4.2×107N/C
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B
5.2×107N/C
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C
6.2×107N/C
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D
7.2×107N/C
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Solution
The correct option is B5.2×107N/C Given: L=10cm=10×10−2m Q=+50μC=+50×10−6C
∴λ=QL=50×10−610×10−2=5×10−4C/m
For the point to be at 10cm from both the ends of the rod, the point must be on the equatorial line.
From the figure,
α=β=θ(let)=30∘
and, r=10cos30∘=5√3cm=5√3×10−2m
From formula, electric field at the equatorial point P