[You do not need to have any prior knowledge of electricity to deal with this passage.]
In a parallel-plate capacitor, two parallel metal plates are connected to a voltage source which maintains a potential V across the plates. Positive charges collect on one side of the capacitor and negative charges on the other side, thus creating an electric field E between the plates. The magnitude of the electric field is related to the potential and the separation between the plates according to
V = Ed
where V is measured in volts, E in J/m, and d in m. A charged particle placed between the plates will experience a force given in magnitude by
F = qE
where q is the charge of the particle in Coulombs, and F is the force in N.
If a new battery is installed, so that the voltage between the plates is increased by a factor of 9, how is the electric field affected?