C Chegg Study I Guided solutio x Homework: Capacitors x C secure https:// /www.f

Question

C Chegg Study I Guided solutio x Homework: Capacitors x C secure https:// /www.flipitphysics.co m/Course/ViewProblem?unitltemID-2442024&enrollmentID-233462; Two parallel plates, each having area A-2695cm2 are connected to the terminals of a battery of voltage Vb-6V as shown. The plates are separated by a distance d 0.38cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate. 1) What is C, the capacitance of this parallel plate capacitor? F Submit 2) What is Q, the charge stored on the top plate of the this capacitor? C Submit 3) A dielectric having dielectric constant K 4.2 is now inserted in between the plates of the capacitor as shown. The dielectric has area A 2695 cm2 and thickness equal to half of the separation (a 0.19 cm What is the charge on the top plate of this capacitor? d/2 uc submit Interactive Example Network Standard Exercise Circuit with Capacitors and a Battery Standard Exercise Parallel Plate Capacitor and Dielectric Standard Exercise Concentric Cylindrical Conducting Shells

Explanation / Answer

part 1:

Capaciatnce C = eo A/d

A is area = 2695 cm^2 = 2695 *10^-4 m^2

d is distance between the plates

eo is permittivity constant = 8.85 *10^-12

so

C = (8.85 *10^-12 * 2695 *10^-4)/(0.0038)

C = 0.000627 uF

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part 2 :

charge Q = CV

Q = 0.000627 *10^-6 * 6

Q = 0.003762 uC

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part 3:

Total Capacitance C = C1+ C2

Ctotal = KeoA/d/2 = eoA/d/

Ctotal = 2eoA/d *(k+1)

Ctotal = 2* 8.85*10^-12 * 2695*10^-4/(0.0038) *(4.2 +1)

Ctotal = 0.00652 uF

Charge Q = CV = 0.00652 *10^-6 * 6

Q = 3.912 *10^-2 uC

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part 4 :

energy Stored U = 0.5 QV

U = 0.5 * 3.912*10^-8 * 6

U = 1.17 *19^-7 Joules

-----------------------------------------

part 5 :

V = Q/C

V = (3.912*10^-8/0.000627 *10^-6)

V = 62.3 Volts

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