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JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

1 Medium

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Application +4 | -1

What is the magnitude of a point charge due to which the electric field 30cm away has the magnitude

2\;newton/coulomb

[1/4\pi {\varepsilon _0} = 9 \times {10^9}N{m^2}/{C^2}]

View answer and explanation

Correct option: A

2 \times {10^{ - 11}}\;coulomb

Option A (

2 \times {10^{ - 11}}\;coulomb

) is correct because it is calculated directly from the formula for the electric field due to a point charge.

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Assertion:- Electric field lines do not cut at any point.
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are separated by a distance

r.

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

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5 \times {10^{ - 2}}

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