Answer
$Q = -1.11 \times 10^{-5} C$
Work Step by Step
To solve this question we need to use two force equations which is $Coulomb$ force, $F = \frac{Qq}{4\pi \epsilon_o r^2}$ and $Centripetal$ force, $F = -\frac{mv^2}{r}$
Equate both equations so that it becomes
$\frac{Qq}{4\pi \epsilon_o r^2} = -\frac{mv^2}{r}$
Here we solve for Q
$Q = -\frac{mv^2 4\pi \epsilon_o r^2}{qr}$
$Q = -\frac{mv^2 4\pi \epsilon_o r}{q}$
Where $4\pi \epsilon_o = \frac{1}{8.99 \times 10^{9} N.m^2/C^2}$,
$q = 4.00 \times 10^{-6} C, $
$m = 0.000800 kg$
$v = 50.0 m/s$
$r = 0.2 m$
$Q = -\frac{(0.0008 kg)(50.0 m/s)^2 (0.2m)}{(8.99 \times 10^{9} N.m^2/C^2)(4.00 \times 10^{-6} C)}$
$Q = -1.11 \times 10^{-5} C$
