Question

Two parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and with plastic in capacitor 2. The potential difference between the plates is the same in both capacitors.a) Compare the magnitudes of the electric fields E1 and E2 between the plates.b) Compare the energies stored in the capacitors U1e and U2e.

Answer 1

Explanation:

The capacitance in two parallel-plate capacitors is:

$C=\frac{K\epsilon_oA}{d}$

For air, we have $K_1=1$

For plastic, we have $K_2=2.25$

Hence:

$C_1=\frac{K_1\epsilon_oA}{d_1}=\frac{\epsilon_oA}{d_1}\\C_2=\frac{K_2\epsilon_oA}{d_2}=2.25(\frac{\epsilon_oA}{d_2})$

a) Recall that the potential difference between the plates is the same ($V_1=V_2=V$). The electric field is given by:

$E_1=\frac{V_1}{d_1}=\frac{V}{d_1}\\E_2=\frac{V_2}{d_2}=\frac{V}{d_1}$

The potential difference is defined as:

$V_1=\frac{Q_1}{C_1}=V\\V_2=\frac{Q_2}{C_2}=V$

Replacing:

$E_1=\frac{Q_1}{C_1d_1}\\E_2=\frac{Q_1}{C_1d_2}\\\\E_1d_1=\frac{Q_1}{C_1}\\E_2d_2=\frac{Q_1}{C_1}\\E_1d_1=E_2d_2\\\frac{E_1}{E_2}=\frac{d_2}{d_1}$

b) The energy is defined as:

$U=\frac{1}{2}CV^2$

So:

$U_1=\frac{1}{2}C_1V_1^2=\frac{1}{2}C_1V^2\\U_2=\frac{1}{2}C_2V_2^2=\frac{1}{2}C_2V^2\\U_1=\frac{1}{2}\frac{\epsilon_oA}{d_1}V^2\\U_2=\frac{1}{2}\frac{2.25\epsilon_oA}{d_2}V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=V^2\\\frac{2U_1d_1}{\epsilon_oA}=V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=\frac{2U_1d_1}{\epsilon_oA}\\\frac{U_1}{U_2}=0.44\frac{d_2}{d_1}$