A parallel plate capacitor is constructed with a dielectric slab with Kappa = 1.

Question

A parallel plate capacitor is constructed with a dielectric slab with Kappa = 1.5 inserted between the plates. The area of each plate is 5 cm^2, and the distance between the two plates is 1 mm. Assume the infinite plane approximation.

1. If we fix the charge on the capacitor to be Q = 6E-11 C, what is the potential difference between the top and bottom plates? Note that in the region containing the dielectric medium, the electric field is E = E0 / Kappa, where E0 is the electric field in a vacuum. (Also, what does it mean by electric field in a vacuum?)

2. What is the capacitance of this configuration? Note: 1 pF = 10^-12 F

Please show complete work and also describe your reasoning and process.

Explanation / Answer

Here ,

1)

as the electric field is given as

E = Q/(area * epsilon)

E = 6 *10^-11/(5 *10^-4 * 1.5 * 8.854 *10^-12)

E = 9035.4 N/C

the electric field in the capacitor is 9035.4 N/

2)

Capacitance , C = Q/V

C = Q/(E * d)

C = 6 *10^-11/(9035.4 * 0.001)

C = 6.641 *10^-12 F

the capacitance for this configuration is 6.641 pF

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