Q · 24

Swapping the dielectric in a charged capacitor

Dielectrics

Question

A charged capacitor when filled with a dielectric \(K = 3\) has charge \(Q_0\), voltage \(V_0\) and field \(E_0\). If the dielectric is replaced with another one having \(K = 9\), the new values of charge voltage and field will be respectively

The Picture — before & after

K = 3 +Q₀ −Q₀ swap K K = 9 +Q₀ −Q₀

“A charged capacitor” = isolated (no battery attached), so the charge \(Q_0\) cannot change. The open dots mark the disconnected terminals.

Given

Initial dielectric\(K_1 = 3\)
Initial values\(Q_0,\; V_0,\; E_0\)
New dielectric\(K_2 = 9\)
Stateisolated ⇒ \(Q\) constant

Asked

New charge\(Q' = ?\)
New voltage\(V' = ?\)
New field\(E' = ?\)

Concept

An isolated charged capacitor keeps its charge fixed. Capacitance scales with the dielectric constant, \(C = K\,\dfrac{\varepsilon_0 A}{d}\), so changing \(K\) changes \(C\) in the same ratio. With \(Q\) held constant:

$$ V=\frac{Q}{C}\propto\frac1K, \qquad E=\frac{V}{d}\propto\frac1K. $$

Method & Steps

  1. Capacitance ratio: \(\dfrac{C'}{C}=\dfrac{K_2}{K_1}=\dfrac{9}{3}=3\Rightarrow C'=3C.\)
  2. Isolated ⇒ charge unchanged: \(Q' = Q_0.\)
  3. New voltage: \(V'=\dfrac{Q_0}{C'}=\dfrac{Q_0}{3C}=\dfrac{V_0}{3}.\)
  4. New field: \(E'=\dfrac{V'}{d}=\dfrac{E_0}{3}.\)

Easy Trick

Isolated capacitor → charge is locked. Whatever factor \(K\) grows by, \(V\) and \(E\) shrink by that same factor. \(K\) went \(\times 3\), so \(V_0\to V_0/3\), \(E_0\to E_0/3\), and \(Q_0\) stays put. (If a battery were connected, \(V\) would be locked and \(Q\) would grow \(\times 3\) instead.)

d
\(Q_0,\;\; \dfrac{V_0}{3},\;\; \dfrac{E_0}{3}\)
Charge fixed (isolated); voltage and field both fall by the factor by which \(K\) rose.