JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

An infinite number of electric charges each equal to

5\;nano - coulomb

(magnitude) are placed along

X

-axis at

x = 1\;cm,\;\;x = 2\;cm,\;\;x = 4\;cm\;\;x = 8\;cm \ldots \ldots \ldots

and so on. In the setup if the consecutive charges have opposite sign, then the electric field in

Newton/Coulomb

X = 0

\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N - {m^2}/{c^2}} \right]

View answer and explanation

Correct option: C

36 \times {10^4}

Option C (

36 \times {10^4}

) is correct because the electric field is calculated by summing the contributions of all charges, considering their alternating signs and positions.

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r.

4N

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

N

5 \times {10^{ - 2}}