## College Physics (4th Edition)

(a) The potential difference between the plates is $0.050~V$ (b) The plate with the positive charge is at a higher potential.
(a) We can find the potential difference between the plates: $\Delta V = \frac{Q}{C}$ $\Delta V = \frac{0.75\times 10^{-6}~C}{15.0\times 10^{-6}~F}$ $\Delta V = 0.050~V$ The potential difference between the plates is $0.050~V$. (b) The electric field points from the positive charges to the negative charges. Since the electric field points from places of higher potential to places of lower potential, the plate with the positive charge is at a higher potential.