Capacitance potential difference Problem

[mustang1988]

1. A 2.05 µF capacitor (#1) is charged to 901 V and a 6.68 µF capacitor (#2) is charged to 671 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each



2. C=8.85e10(A/d), Q=v/c,

The Attempt at a Solution

Im not really sure where to start, i think that in the end both voltages should be the same but not sure how to get it there. Any help would be great, thanks.

 

[rl.bhat]

Find charge in each capacitor using the relevant formula. When the positive plates ate connected, the total charge in the combination is the sum of the charges and the combination is the parallel combination. Find the capacitor of the combination and find the voltage across the combination.

 

[mustang1988]

Ok i get the part about adding the charges but when you say find the capacitor of the combination and find the voltage across the combination I am confused.

 

[rl.bhat]

Ok i get the part about adding the charges but when you say find the capacitor of the combination and find the voltage across the combination I am confused.

In the parallel combination C = C1 + C2.
Common V = (Q1 + Q2)/(C1 + C2)

 

[mustang1988]

ok that made more sense thanks a lot

 

[mustang1988]

i have one more question.
The problem is: If a capacitor has opposite 4.0 µC charges on the plates, and an electric field of 2.6 kV/mm is desired between the plates, what must each plate's area be?

Im pretty sure the Equation is E=(Q/A)(4pik) but i can't figure out how to get 2.6kV/mm into m^2, which is what they want the answer in

 

[drizzle]

i have one more question.
The problem is: If a capacitor has opposite 4.0 µC charges on the plates, and an electric field of 2.6 kV/mm is desired between the plates, what must each plate's area be?

Im pretty sure the Equation is E=(Q/A)(4pik) but i can't figure out how to get 2.6kV/mm into m^2, which is what they want the answer in

first convert the units, then solve
[kV/mm]→ [kV/(1/1000)m]→ [1000kV/m]
k→ 1000
μ→ 10^-6

 

[mustang1988]

Once i solve for A will that be in m^2?

 

[rl.bhat]

i have one more question.
The problem is: If a capacitor has opposite 4.0 µC charges on the plates, and an electric field of 2.6 kV/mm is desired between the plates, what must each plate's area be?

Im pretty sure the Equation is E=(Q/A)(4pik) but i can't figure out how to get 2.6kV/mm into m^2, which is what they want the answer in

From where did you get E=(Q/A)(4pik) formula?
Actual formula is
C = ε_οΑ/d
Now between the charged parallel plates E = V/d = Q/Cd
Or C = Q/Ed = ε_οΑ/d
Or A = Q/E*ε_o