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Question

A 10 cm long rod carries a charge of +50μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both the ends of the rod.

A
4.2×107 N/C
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B
5.2×107 N/C
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C
6.2×107 N/C
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D
7.2×107 N/C
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Solution

The correct option is B 5.2×107 N/C
Given:
L=10 cm=10×102 m
Q=+50μC=+50×106 C

λ=QL=50×10610×102=5×104 C/m

For the point to be at 10 cm 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=53 cm=53×102 m

From formula, electric field at the equatorial point P

EP=2kQsinθLr=2kλsinθr

EP=2×9×109×5×104×sin3053×102

EP=5.2×107 N/C

Hence, option (b) is the correct answer.

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