Two 15-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The

Question

Two 15-cm-diameter electrodes

0.40 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 12 V battery.

Part A

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?

Part D

What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.80 cm apart? The electrodes remain connected to the battery during this process.

Express your answer to two significant figures and include the appropriate units.
Q=?
E=?
DeltaV=?

Part G

What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of parts D-F) are expanded until they are 30 cm in diameter while remaining connected to the battery?

Express your answer to two significant figures and include the appropriate units.

Q=?
E=?
DeltaV=?

Explanation / Answer

you have merged 3 parts in part a) so I am answering only 3 sub parts.i.e part a)

a) q1 = C*v
where C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.075)^2/0.0040 = 3.909x10^-11F

So q1 = 3.909x10^-11F*12 = 4.6908x10^-10C and q2 = -4.6908x10^-10C

(b) E = (V / d) = (12 / 0.0040) = 3000 V/m

c) V = E*d = 3000V/m*0.0040 = 12.0V

raise another point for d,g,f sub parts.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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