SECTION - A The magnitude of the electric field due to a point charge object at a distance of 4.0 {\m} is 9 { {\N}}/{ {\C}}. From the same charged object the electric field of magnitude, 16 { {\N}}/{ {\C}} will be at a distance of

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 {\m}

9 \;{ {\N}}/{ {\C}}

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

16 \;{ {\N}}/{ {\C}}

will be at a distance of

Solution:

1^{\;\text "st "\;}

case,

\;E=\;{k q}/{r^2}

\;\;\text " Or, "\; \;\; 9=\;{k q}/{4^2} \;\; [r=4.0 {\m}]

\; ∴ \;\; k q=9 × 16

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

\;E^{'}=\;{k q}/{ (r^{'}) ^2}

\;\text " Or, "\;\;16\;=\;{9 × 16}/{ (r^{'}) ^2}

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

∴ \;r^{'}\;=3 {\m}

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