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1. Electric Charges and Fields
easy
Two positive charges of $20$ $coulomb$ and $Q\;coulomb$ are situated at a distance of $60\,cm$. The neutral point between them is at a distance of $20\,cm$ from the $20\,coulomb$ charge. Charge $Q$ is.....$C$
A
$30$
B
$40$
C
$60$
D
$80$
Solution
(d) At neutral point
$k \times \frac{{20}}{{{{(20 \times {{10}^{ – 2}})}^2}}} = k \times \frac{Q}{{{{(40 \times {{10}^{ – 2}})}^2}}}$ $==>$ $Q = 80 \,C$
Standard 12
Physics