## SAT Physics Subject Test

**Chapter 9 ****Electric Potential and Capacitance**

**DIELECTRICS**

One method of keeping the plates of a capacitor apart, which is necessary to maintain charge separation and store potential energy, is to insert an insulator (called a **dielectric**) between the plates.

A dielectric always increases the capacitance of a capacitor.

Let’s see why this is true. Imagine charging a capacitor to a potential difference of Δ *V* with charge +*Q* on one plate and –*Q* on the other. Now disconnect the capacitor from the charging source and insert a dielectric. What happens? Although the dielectric is not a conductor, the electric field that existed between the plates causes the molecules within the dielectric material to polarize; there is more electron density on the side of the molecule near the positive plate.

The effect of this is to form a layer of negative charge along the top surface of the dielectric and a layer of positive charge along the bottom surface; this separation of charge induces its own electric field (**E**_{i}), within the dielectric, which opposes the original electric field, **E**, within the capacitor.

So the overall electric field has been reduced from its previous value: **E**_{total} = **E** + **E**_{i}, and **E**_{total}= **E** – **E**_{i}. 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

E_{with dielectric} = E_{without dielectric} –E_{i} =

Since ∆*V* = E*d* for a parallel-plate capacitor, we see that ∆*V* must have decreased by a factor of k. But *C* = , so if ∆*V* decreases by a factor of k, then *C* increases by a factor of k:

*C*_{with dielectric} =k*C*_{without dielectric}

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

9. 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

*C*_{with dielectic} = k*C*_{without 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.