Please provide step by step solution Two parallel plates, each having area A- 38

Question

Please provide step by step solution

Two parallel plates, each having area A- 3834cm are connected to the terminals of a battery of voltage Vb 6 V as shown. The plates are separated by a distance d = 0.43cm. 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? 0.000789 F Submit You currently have 3 submissions for this question. Only 6 submission are allowed. You can make 3 more submissions for this question. Your submissions: 0.000789 Computed value: 0.000789 Feedback: Your answer has been judged correct, the exact answer is: 7.88372825876869 Submitted: Monday, March 28 at 6:37 PM 10.4. 2) What is Q, the charge stored on the top plate of the this capacitor? 0.00473 Submit You currently have 2 submissions for this question. Only 6 submission are allowed. You can make 4 more submissions for this question Your submissions: 0.00473 Computed value: 0.00473 Feedback: Your answer has been judged correct; the exact answer is: 0.00473023695526121. Submitted: Monday, March 28 at 6:39 PM

Explanation / Answer

parallel plate capacitor of area A = 3834 cm2
battery voltage Vb = 6 V
d = 0.43 cm
1)
now capacitance C = epsilon not *A/d = 8.85*10^-12*3834*10^-4/0.43*10^-2 = 7.8909069767442*10^-10 F= 0.000789 micro farad

2)
charge Q = C*V = 7.8909069767442*10^-10*6 = 4.7345441860465*10^-9 C = 0.004734 micro coulombs

when a dielectric materia of constant k is inserted with half the width between the plates then

the capacitance of the upper plate is as (per given problem ) C = epsilon not *A/d/2 = 2*C= 2*0.000789*10^-6 F = 1.578*10^-9 F

3)
   now charge on the top plate of capacitor is Q = C*V = 2*0.000789*10^-6 *6 = 9.468*10^-9 C

4)

    energy stored in the capacitor U = 1/2 cV^2 = 0.5*1.578*10^-9 *6^2 = 2.8404*10^-8 J

5) Q = CV ==> v=Q/C = 9.468*10^-9 / 0.000789 *10^-6 = 12 V

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