Two parallel plates, each having area A = 2044 cm^2 are connected to the termina

Question

Two parallel plates, each having area A = 2044 cm^2 are connected to the terminals of a battery of voltage V_b = 6 V as shown. The plates are separated by a distance d = 0.39 cm. What is Q. the charge on the top plate? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is U, the energy steed in the this capacitor? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. The battery is now disconnected from the plates and the separation of the plates is doubled (- 0.78 cm). What is the energy stored in this new capacitor? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is E. the magnitude of the electric field in the region between the plates? You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. Compare V, the magnitude of the new potential difference across the plates, to V_b, the voltage of the battery. V V_b You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Explanation / Answer

Given

capacitor of area A = 2044 cm2 = 0.2044 m2

the potential difference is Vb = 6 V

separation of the plates is d = 0.0039 m

the capacitance of the capacitor is C = epsilon not *A/d

   C = 8.854*10^-12*0.2044 /(0.0039) F

   C = 4.6404041025641*10^-10 F

1)   C = 464.04041 pF

Q = C*V

Q = 464.04041 *10^-12*6 C = 2.78424246*10^-9 C = 2.78424246 nC

2) energy stored in the capacitor i u = 0.5*C*V^2

   U = 0.5*464.04041*10^-9*6^2 J = 8.35272738*10^-6 J

3) energy stored is u = q^2/2C = 0.5*C*V^2

and C = epsilon not*A/d

   C' = C/2

energy stored is U' = 0.5(C/2)v^2

       = U/2

       U' = 8.35272738*10^-6/2 = 4.17636369*10^-6 J

4) electric field between the plates is

   E = sigma / epsilon not

where sigma is surface charge density

   sigma = q/area = 2.78424246*10^-9 /(0.2044) C/m2 = 1.362154*10^-8 C/m2

now the field is E = 1.362154*10^-8 /(8.854*10^-12) N/C
      
       E = 1538.462 N/C

5) now potential difference is V = E*d

           v = 1538.462*0.0039 V = 6.0000018 V

v > vb

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