Question

Problem 2: A parallel plate capacitor has a charge on one plate of q= 8.5x10^-7 C. Each square plate is d1= 1.1 cm wide and the plates of the capacitor are separated by d2= 0.35mm. The gap is filled with air, E0= 8.85x10^-12 C^2/Nm^2

Part A) What is the voltage between the plates, deltaV, in V?

Cassandra Chapter 19-part 2 (19b) Begin Date: 1/22/2019 1:00.00 AM-Due Date: 1/27/2019 11:55:00 PM End Date: 128/2019 11:55.00 PM (20%) Problem 2: A parallel plate capacitor has a charge on one plate of 9-8.55-07 С Each square plate is d-1,1 cm wide and the plates of the capacitor are separated by d 0.35 mm. The gap is filled with air, 885 10-12 CNm 50% Part (a) What is the voltage between the plates, ar. Grade Sammary Deductions Potesnal 100 coso depts resaing acos) acotan) sitho asinO cosh . Degrees Radis Sunday 1/27/2079 acer

Answer 1

Capacitance = e_0A/d_2 = 8.85 times 10^-12 times d_1^2/0.35 times 10^-3 = 8.85 times 10^-12 times (1.1 times 10^-2)^2/0.35 times 10^-3 = 3.06 times 10^-12 F voltage = q/c = 8.5 times 10^-7/3.06 times 10^-12 V = 2.78 times 10^5 V now voltage (v') = 2v = 2q/e_0 A/d [C = e_0 A/d] = qd/e_0 A/2 For A/2 = (d_1/squareroot 2)^2 that means the width of plate should be reduced to 1/squareroot 2 of original so, d'_1 = d_1/squareroot 2 = 1.1 times 10^-2/squareroot 2 = 7.78 times 10^-3 m d'_1 = 7.78 mm