Answer
$5\times10^{-9}C$
Work Step by Step
Because of the problem’s spherical symmetry, outside the ball/pea the electric field strength can be calculated using equation 16–4a. Solve for the charge.
$$E=k\frac{Q}{r^2}$$
$$Q=\frac{Er^2}{k}$$
$$Q=\frac{(3\times10^6N/C)(0.00375m)^2}{8.988\times10^{9}(N \cdot m^2)/C^2}$$
$$Q=5\times10^{-9}C $$
This is the charge of about 30 billion electrons.
