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1. Electric Charges and Fields
easy
What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$
A
$8 \times {10^{ - 11}}\,C$
B
$2 \times {10^{ - 12}}\,C$
C
$3 \times {10^{ - 11}}\,C$
D
$6 \times {10^{ - 10}}\,C$
Solution
(a) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{{{r^2}}}$ $==>$ $2 = 9 \times {10^9} \times \frac{Q}{{{{(0.6)}^2}}}$ $==>$ $Q = 8 \times {10^{ – 11}}\,C$
Standard 12
Physics