Cracking the SAT Physics Subject Test

So the overall electric field has been reduced from its previous value: Etotal = E +

Ei, and Etotal= E − Ei. Let’s say that the electric field has been reduced by a factor

of k (the Greek letter kappa) from its original value as follows

Since ∆V = Ed for a parallel-plate capacitor, we see that ∆V must have decreased

by a factor of κ. But C = , so if ∆V decreases by a factor of κ, then C
increases by a factor of κ:

Cwith dielectric = κCwithout dielectric

The value of κ, called the dielectric constant, varies from material to material, but
it’s always greater than 1.

8. A parallel-plate capacitor with air between its plates has a
   capacitance of 2 × 10−6 F. What will be the capacitance if the
   capacitor is fitted with a dielectric whose dielectric constant is 3?

Here’s How to Crack It

The capacitance with a dielectric is equal to k times the capacitance without, where
k is the dielectric constant. So

Cwith dielectic = κCwithout dielectic = (3)(2 × 10−6 F) = 6 × 10−6 F.

Remember that a capacitor with a dielectric always has a greater capacitance than
the same capacitor without a dielectric.