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JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

A circular loop of radius a carries a total charge

Q

distributed uniformly over its circumference. A small length

{dL}

of the wire is cut off. The magnitude of electric field at the center due to remaining wire will be

\left[K=\frac{1}{4 \pi \varepsilon_0}\right]

View answer and explanation

Correct option: C

\frac{K Q d L}{2 \pi a^3}

Option D (

\frac{{K Q d L}}{{a^3}}

) is incorrect. The correct answer is C (

\frac{{K Q d L}}{{2 \pi a^3}}

) because it calculates the electric field due to the missing charge element.

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\left( {{\varepsilon _r} = 8} \right)

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