Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 13 V battery.

Part A. What is the charge on each electrode while the capacitor is attached to the battery?

Part B. What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

Part C. What is the potential difference between the electrodes while the capacitor is attached to the battery?

Part D. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.

Part E. What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.

Part F. What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

Given

electrodes fo diameter is d = 13cm, radius r = 0.065 m

separated by a distance d = 0.59 cm = 0.0059 m

   connected to a battery of e = 13 V

we know that the capacitance of a prallel plate capacitor is C = epsilon0*a/d

       C = 8.854*10^-12*(pi*0.065^2)/(0.0059) F

       C = 1.991884*10^-11 F

       C = 19.91884 pF

a) charge on the each electrode is Q = c*V = 19.91884*10^-12*13 C = 2.5894492*10^-10 C

b)the electric field strength inside the capacitor is E = sigma /epsilon0

       E = (q/A*epsilon0)
       E = ((19.91884*10^-12) /(pi*0.065^2)(8.854*10^-12)) N/C

       E = 169.49150 N/C

c) potential differnce between the electrodeds V = Q/C = 2.5894492*10^-10/(19.91884*10^-12) V = 13 V

d)if the electrodes are separated by a distance of d = 1.7 cm = 0.017 m

the capacitance of a prallel plate capacitor is C = epsilon0*a/d

       C = 8.854*10^-12*(pi*0.065^2)/(0.017) F

       C = 6.9130*10^-12 F

       C = 6.9130 pF

charge on the each electrode is Q is unchanged = 2.5894492*10^-10 C

e)
  

       E = (q/A*epsilon0)

       E = ((8.9869*10^-11) /((pi*0.065^2)(8.854*10^-12))) N/C

       E = 764.705 N/C

f) potential difference is V = Q/C = (2.5894492*10^-10) /(6.9130*10^-12) V = 37.46 V

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.