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1. Electric Charges and Fields
easy
$ + 2\,C$ and $ + 6\,C$ two charges are repelling each other with a force of $12\,N$. If each charge is given $ - 2\,C$ of charge, then the value of the force will be
A
$4\,N$ (Attractive)
B
$4\,N$ (Repulsive)
C
$8\,N$ (Repulsive)
D
Zero
Solution
(d) Resultant charges after adding the $-2\,C$ be $( – \,2 + 2) = 0$ and $( – \,2 + 6) = + \,4\,C$ $==> F$ $ = \frac{{k\,{Q_1}{Q_2}}}{{{r^2}}} = k \times \frac{{0 \times 4}}{{{r^2}}} = 0$
Standard 12
Physics
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