Question

Two charges of +25 μC and +16 μC are 90 mm apart. A third charge of -60 μC is placed on a line joining the two charges, 30 mm from the +25 μC charge. Find the net force on the third charge.

Answer 1

Answer

net force on the third charge

Is due to all other charge is

$F_1=\frac{kqQ}{r^2}\\=\frac{9*10^9*25*60*10^{-12}}{900*10^{-6}}\\=15*10^3N$

Due to other charge

$F_2=\frac{kq'Q}{r^2}\\=\frac{9*10^9*16*60*10^{-12}}{3600*10^{-6}}\\=2.4*10^3N$

So net force on third charge

F=15000-2400=12600N