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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.

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Explanation:

The capacitance in two parallel-plate capacitors is:

[tex]C=\frac{K\epsilon_oA}{d}[/tex]

For air, we have [tex]K_1=1[/tex]

For plastic, we have [tex]K_2=2.25[/tex]

Hence:

[tex]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})[/tex]

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

[tex]E_1=\frac{V_1}{d_1}=\frac{V}{d_1}\\E_2=\frac{V_2}{d_2}=\frac{V}{d_1}[/tex]

The potential difference is defined as:

[tex]V_1=\frac{Q_1}{C_1}=V\\V_2=\frac{Q_2}{C_2}=V[/tex]

Replacing:

[tex]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}[/tex]

b) The energy is defined as:

[tex]U=\frac{1}{2}CV^2[/tex]

So:

[tex]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}[/tex]

onyebuchinnaji

(a) The magnitudes of the electric fields  between the plates is [tex]E_1 = 2.25(\frac{E_2C_1}{C_2} )[/tex].

(b) The energy stored in the capacitors when compared is [tex]U_1 = U_2 (\frac{C_1}{C_2} )[/tex]

The capacitance of the two capacitor is given as;

[tex]C= \frac{k\epsilon _0 \ A}{d}[/tex]

where;

  • k is dielectric constant, air = 1 and plastic = 2.25

The capacitance for the first capacitor;

[tex]C_1= \frac{1 \times \epsilon _0 \ A}{d}= \frac{ \epsilon _0 \ A}{d_1}\\\\d_1 = \frac{\epsilon _0 \ A}{C_1}[/tex]

The capacitance of the second capacitor;

[tex]C_2= \frac{2.25 \times \epsilon _0 \ A}{d_2}\\\\d_2 = \frac{2.25 \times \epsilon _0 \ A}{C_2}[/tex]

The potential difference and electric field between the plates is given as;

[tex]Q = CV\\\\V = \frac{Q}{C} \\\\E = \frac{V}{d} \\\\E = \frac{Q}{Cd}\\\\Ed = \frac{Q}{C} = V\\\\E_1d_1 = E_2 d_2\\\\E_1 = \frac{E_2d_2}{d_1} \\\\E_1 = E_2(\frac{d_2}{d_1})\\\\E_1 = E_2 (\frac{2.25\times \epsilon _0 A}{C_2} \times \frac{C_1}{\epsilon _0 A} )\\\\E_1 = E_2(\frac{2.25C_1}{C_2} )\\\\E_1 = 2.25(\frac{E_2 C_1}{C_2} )[/tex]

(b)

The energy stored in each capacitor is given as;

[tex]U_1 = \frac{1}{2} C_1 V^2\\\\U_2 = \frac{1}{2} C_2 V^2\\\\\frac{U_1}{U_2} = \frac{2C_1V^2}{2C_2V^2} \\\\\frac{U_1}{U_2} = \frac{C_1}{C_2}[/tex]

[tex]U_1 = U_2 (\frac{C_1}{C_2} )[/tex]

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