Capacitance Homework Word Problem

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
mer584
Messages
37
Reaction score
0
The problem.
A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. 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? (note that charge is conserved)


Homework Equations


PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va


Attempt:
What I started problem I used the formula Q=CV to find the charge that is occurring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform field so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?
 
Physics news on Phys.org
I would compute the capacitance of the two capacitors in parallel, then use Q=CV for this capacitor to get the potential difference across both, and then use Q=CV for the 2 original capacitors to get the charge on each