First slide

Electrostatic force

Question

Two charges each equal to $2 μC$ are 0.5 m apart. If both of them exist inside vacuum, then the force between them is

Easy

A

1.89 N
B

2.44 N
C

0.144 N
D

3.144 N

Solution

By using $F = 9 \times 10^{9} \cdot \frac{Q^{2}}{r^{2}}$
$\Rightarrow F = 9 \times 10^{9} \cdot \frac{{(2 \times 10^{- 6})}^{2}}{(0 .5)^{2}} = 0 .144 N$

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