Bb Bb Bb IU 68%_ 2:40 PM Problem 18.35 MasteringPhysics Conslanls You make a cap

Question

Bb Bb Bb IU 68%_ 2:40 PM Problem 18.35 MasteringPhysics Conslanls You make a capacitor by cutting the 15.5cm diameter bottoms out of two aluminum pie plates, separating them by 3.35 mm, and connecting them acrass a 6.00 V battery Part A What's the capacitance of your capacitor? Express your answer to three slgnfficant figures with the appropriate units. C-49.9 ph Correct Part B If you discannect the battery and separate the plates to a distance of 3.50 CT without discharging lhem, whal will be lhe polenlial difference between them? Express your answer to three significant figures with the appropriate units. 6.27 V SubmitRequest Answer Incorrect; Try Again; 2 attempts remaining

Explanation / Answer

first of all charge will remain same as in part 1

Q = C*V

Q = 49.9*10^-12*6

Q = 3.0*10^-10 C

Now, We know that

C = e0*A/d

Now if distance is increased, then new capacitance will be

C = 8.85*10^-12*pi*0.155^2/(4*3.50*10^-2)

C = 4.77*10^-12 F = 4.77 pF

Now new potential difference will be

V = Q/C

V = 3.00*10^-10/(4.77*10^-12)

V = 62.893 V = 62.9 V

Please Upvote.

Now I think your mistake was that distance is increased to 3.35 mm to 3.50 cm. Notice the change in units.

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