Question

Two positive charges of magnitude 20 micro C and 100 micro C are placed at a distance of 150cm. Calculate the force of repulsion between them.

Answer 1

$first \: positive \: charge = q1 = 20$

$1micro \: = {1}^{ - 6}$

$1q = 20 \times 10 {}^{ - 6}$

$second \: charge = q2 = 100 \times 10 {}^{ - 6}$

$formula = f = \frac{k \times q1 × q2}{d {}^{2} }$

remember

$k = 9 \times {10}^{9}$

change distance 150cm to 1.5m

putting values

f =$\frac{9 \times 10 {}^{9} \times 20 \times 10 {}^{ - 6} \times 100 \times {10}^{ - 6} }{1.5 {}^{2} }$

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answer

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