Physics - Center of Mass Question with Solution | TestHub

For each ball: $m_{2}g-T-q_{2}E=m_{2}a$ and $T+q_{1}E-m_{1}g=m_{1}a$.

Gravity and the electric force on the whole system cancel, but the pulley exerts an external force 2T on the system, and solving gives $T=\frac{(m_1-m_2)}{2}a$.

$a_{cm}=\frac{2T}{m_1+m_2}\ne 0$ in general (zero only if $m_1=m_2$). So it is not necessarily zero. Hence (B).