CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

Question : 1 of 35

Marks: +1, -0

SECTION - A

The magnitude of the electric field due to a point charge object at a distance of

4.0\,\text{m}

9\,\frac{\mathrm{N}}{\mathrm{C}}

. From the same charged object the electric field of magnitude,

16\,\frac{\mathrm{N}}{\mathrm{C}}

will be at a distance of

Solution: 👈: Video Solution

1^{\text{st}}

case,

E=\frac{kq}{r^2}

\text{Or, }9=\frac{kq}{4^2}\,[r=4.0\,\text{m}]

\therefore kq=9\times 16

\text{In }2^{\text{nd}}\text{ case,}

E'=\frac{kq}{(r')^2}

\text{Or, }16=\frac{9\times 16}{(r')^2}

\text{Or, }(r')^2=9

\therefore r'=3\,\text{m}

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