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1. Electric Charges and Fields
medium
A charged particle of mass $0.003\, gm$ is held stationary in space by placing it in a downward direction of electric field of $6 \times {10^4}\,N/C$. Then the magnitude of the charge is
A
$5 \times {10^{ - 4}}\,C$
B
$5 \times {10^{ - 10}}\,C$
C
$ - 18 \times {10^{ - 6}}\,C$
D
$ - 5 \times {10^{ - 9}}\,C$
Solution
(b) By using $QE = mg$
$==>$ $Q = \frac{{mg}}{E} = \frac{{0.003 \times {{10}^{ – 3}} \times 10}}{{6 \times {{10}^4}}}= 5 ×10^{-10} \,C$
Standard 12
Physics
