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1. Electric Charges and Fields
easy
What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N{m^2}/{C^2}]$
A
$2 \times {10^{ - 11}}\,coulomb$
B
$3 \times {10^{ - 11}}\,coulomb$
C
$5 \times {10^{ - 11}}\,coulomb$
D
$9 \times {10^{ - 11}}\,coulomb$
Solution
(a) Electric field due to a point charge $E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}$
$q = E \times 4\pi {\varepsilon _0}{r^2} = 2 \times \frac{1}{{9 \times {{10}^9}}} \times {\left( {\frac{{30}}{{100}}} \right)^2}$= $2 \times {10^{ – 11}}\,coulomb$
Standard 12
Physics
