Question

two identical tiny metal balls carry charges 3 microcoulomb and -12 microcoulomb and are made to touch each other. then they are separated to 3cm apart. then the force between them is

Answer 1

Charges before sharing,

Q1 = 3μC, Q2 = -12μC

Charges after sharing,

Q = (Q1+Q2) ÷ 2

Q = (3-12)÷2

Q = 4.5μC

$f = \frac{1}{4\piε} \times \frac{q \times q}{r {}^{2} }$

$f = 9 \times {10}^{9} \times \frac{ - 4.5 \times {10}^{ - 6} \times - 4.5 \times {10}^{ - 6} }{9 \times 10 ^{ - 4} }$

$f =20.25 \times {10}^{ 1}$

Therefore, F = 202.5N