A parallel plate capacitor has the following attributes. The area of each plate

Question

A parallel plate capacitor has the following attributes. The area of each plate is 0.50 m^2 . The distance between the plates is 1.00 × 10^-6 m. The material between the plates is porcelain (Porcelain has a dielectric constant of r = 6.5). The voltage of one plate with respect to the other is 2.0 V.

1) What is the capacitance of the capacitor?

2) What is the electric field strength between the plates of the capacitor?

3) Following the last question: When the material between the plates is replaced with air (r = 1.0), and the plates are disconnected from the charging battery. What will be the voltage between the two plates?

Explanation / Answer

As given in the question,

A = 0.5 m^2, d = 1*10^-6 m,   r = 6.5, V = 2 V

(1)   The capacitance of the capacitor,

    C = r*o*(A/d)   , where   o = 8.854*10^-12 F*m^-1

= 6.5*8.854*10^-12*(0.5 / 1*10^-6) = 2.878*10^-5 F

(2)   The electric field strength between the plates of the capacitor,

  E = V/d = 2 / 1*10^-6 = 2*10^6 V

(3)   The charge stored in the capacitor after first condition,

Q = C*V = 2.878*10^-5 * 2 = 5.756*10^5 C

When r' = 1,

The capacitance of the capacitor,

    C' = (r')*o*(A/d)   , where   o = 8.854*10^-12 F*m^-1

= 1*8.854*10^-12*(0.5 / 1*10^-6) = 4.427*10^-6 F

Now, the voltage between the paltes,

V = Q / C' = 5.756*10^5 / 4.427*10^-6 = 1.3*10^11 V

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