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1. Electric Charges and Fields
easy
Two charges each of $1\;coulomb$ are at a distance $1\,km$ apart, the force between them is
A
$9 \times {10^3}\;Newton$
B
$9 \times {10^{ - 3}}\;Newton$
C
$1.1 \times {10^{ - 4}}\;Newton$
D
${10^4}\;Newton$
Solution
(a) $F = \frac{{k{Q^2}}}{{{r^2}}} = 9 \times {10^9} \times {1^2} \times \frac{1}{{{{(1000)}^2}}} = 9 \times {10^3}N$
Standard 12
Physics
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