## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$q_A = -8.0\times 10^{-8}~C$ $q_B = -8.0\times 10^{-8}~C$
Since the metal spheres are in contact, the electrons will spread around both spheres after the electrons are added to sphere A. Since the spheres are identical, half of the electrons will be on sphere A and half of the electrons will be on sphere B. We can find the charge on sphere A: $q_A = (\frac{1.0\times 10^{12}}{2})(-1.6\times 10^{-19}~C)$ $q_A = -8.0\times 10^{-8}~C$ Since the charge on sphere B is the same as the charge on sphere A, $q_B = -8.0\times 10^{-8}~C$.