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 * 10 {}^( - 6)`

`second \: charge = q2 = 100 * 10 {}^( - 6)`

`formula = f = \frac{k * q1 × q2}{d {}^(2) }`

remember

`k = 9 * {10}^(9)`

change distance 150cm to 1.5m

putting values

f =
`\frac{9 * 10 {}^(9) * 20 * 10 {}^( - 6) * 100 * {10}^( - 6) }{1.5 {}^(2) }`

answer

8N