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1. Electric Charges and Fields
easy
The charge on $500\,cc$ of water due to protons will be
A
$6.0 \times {10^{27}}\,C$
B
$2.67 \times {10^7}\,C$
C
$6 \times {10^{23}}\,C$
D
$1.67 \times {10^{23}}\,C$
Solution
(b) $Q = ne$; where $n$ = number of moles $× 6.02 × 10^{23}× 10$
$==>$ $Q = \frac{{500}}{{18}} \times 6.02 \times {10^{23}} \times 10 \times 1.6 \times {10^{ – 19}} = 2.67 \times {10^7}\,C$
Standard 12
Physics
