A parallel-plate capacitor with circular plates and a capacitance of 11.0 F is c

Question

A parallel-plate capacitor with circular plates and a capacitance of 11.0 F is connected to a battery which provides a voltage of 12.6 V .

Part A

What is the charge on each plate?

Part B

How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery?

Part C

How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation?

Part C

How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation?

Explanation / Answer

As we know that

A parallel-plate capacitor with circular plates and a capacitance = C = 11 uF

Voltage = V = 12.6 V.

Charge = q = CV =11*10^-6*12.6

The charge on each plate =1.386*10^-4 C

B) If their separation were doubled ,capacity is halved but voltage remains same as the capacitor remained connected to the battery
The charge on each plate is halved

The charge on each plate =6.93*10^-5 C

C) If the capacitor were connected to the battery after the radius of each plate was doubled without changing their separation, capacity is four times
The charge on each plate =5.54*10^-4 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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