Two surfaces A and B are enclosing the charge | vedclass

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(C) The total normal electric induction ($T$.$N$.$E$.$I$.) through a closed surface is equal to the algebraic sum of the charges enclosed by that surf

Vedclass · Accepted Answer

$+q$ and $+2 q$

Vedclass · Answer

(C) The total normal electric induction ($T$.$N$.$E$.$I$.) through a closed surface is equal to the algebraic sum of the charges enclosed by that surface.$	ext{T.N.E.I.} = \sum q_{	ext{enclosed}}$For surface $A$,the enclosed charges are $+2q$ and $-q$.$	ext{T.N.E.I. for } A = (+2q) + (-q) = +q$For surface $B$,the enclosed charges are $+3q$ and $-q$.$	ext{T.N.E.I. for } B = (+3q) + (-q) = +2q$Therefore,the $T$.$N$.$E$.$I$. through surfaces $A$ and $B$ are $+q$ and $+2q$ respectively.

Two surfaces $A$ and $B$ are enclosing the charges as shown below. The total normal electric induction ($T$.$N$.$E$.$I$) through the surfaces $A$ and $B$ are respectively.

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