In Figure , two parallel-plate capacitors A and B are connected in parallel acro

Question

In Figure , two parallel-plate capacitors A and B are connected in parallel across a 690 V battery. Each plate has area 67.0 cm2; the plate separations are 4.30 mm. Capacitor A is filled with air; capacitor B is filled with a dielectric of dielectric constant = 5.10. Find the magnitude (in N/C) of the electric field within (a) the dielectric of capacitor B and (b) the air of capacitor A. What are the free charge densities on the higher-potential plate of (c) capacitor A and (d) capacitor B? (e) What is the induced charge density ' on the top surface of the dielectric?

Please answer with correct units and all parts of this problem

Explanation / Answer

Capacitance A = A/d = 1.377 * 10^ -11 F

Capacitance B = KA/d = 7.02 * 10^-11 F

Potential difference across both the capacitors = 690 V
Electric field calculation = V/r

a)Electric field = V/r = 690/.0043 = 160465 N/C

b)Electric field = V/r = 690/.0043 = 160465 N/C

c)Charge at capacitor A = C1V = 1.377 * 10^ -11 * 690 = 9.501 * 10^-9 C

A = 67 * 10^-4 m^2

charge density = 1.41 * 10^-6 C/m^2

d)Charge at capacitor B = C2V = 7.02 * 10^-11 * 690 = 4.8438 * 10^-8 C

A = 67 * 10^-4

Charge density = 4.8438 * 10^-8/67 * 10^-4 = 7.22 * 10^-6 C/m^2

e)Induced charge density in the dielectric will be same as that induced in the capacitor B but with negative charge

i.e = - 7.22 * 10^-6 C/m^2

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