If you could explain as best you can that’d be great, not just looking for answe

Question

If you could explain as best you can that’d be great, not just looking for answers. Two 11-cm-diameter electrodes 0.42 cm apart form a parallel-plate capacitor. The electrodes are attached metal wires to the terminals of a 20 V battery Part A by What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes while the capacitor is attached to the battery? Express your answer to two significant figures and include the appropriate units. Value Units

Explanation / Answer

A) Q = CV = Ae0/d *V = pi*(0.11/2)^2*8.85e-12 /0.0042*20

= 4.0*10^-10 C

B) E = V/d =20/0.0042 = 4.8*10^3 V/m

C) delta V = 20V

D) charge = CV and C becomes half,

Q=2.0*10^-10 C

E)since d becomes twice, E=V/d = 2.4*10^3 V/m

F) delta V =20V

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