Question

The electrostatic force on a small sphere charge 0.4 uc due to another small sphere of charge -0. 8 uc in air is 0.2 n

Answer 1

Explanation:

though you didn't give that what we have to find...but i supposed that we have to find distance between the charges... pls check the ans in the above attachement...it will helpful for u i hope so...thnks for asking the ques..

Answer 2

(a)

we know 1μC = $\tt{10^{-6} }$C

$\frak{Given}\begin{cases}\tt{F=0.2\:N}\\ \ \tt{q_{1}=0.4\times10^{-6}\:C}\\ \ \tt{q_{2}=-0.8\times 10^{-6}\:C}\end{cases}$

$\implies \: \sf{F = \frac{q _{1} \times q _{2}}{4 \pi \epsilon _{0}r {}^{2} }} \\ \\ \\ \longrightarrow \tt{ \frac{0.4 \times 10 {}^{ - 6} \times 8 \times 10 {}^{ - 6} \times 9 \times 10 {}^{9} }{0.2} } \\ \\ \\ \longrightarrow \: \tt{144 \times 10 {}^{ - 4} } \\ \\ \\ \longrightarrow \: \tt {\: r = \sqrt{144 \times 10 {}^{ - 4} } } \\ \\ \\ \longrightarrow \: \boxed{ \bf{0.12 \: m}}$

(b)

Both the spheres attracts each other with the same force. Therefore the force on the second sphere due to the first is 0.2 N.