Potential Difference V_ad: Solving with Switch S Closed

[Black Armadillo]

Homework Statement

The capacitors in the Figure are initially uncharged and are connected, as in the diagram, with switch S open. The applied potential difference is V_ab = + 210V

What is the potential difference V_ad after switch S is closed?

Homework Equations

Q=C_eq*V
V_ab=Q/C

The Attempt at a Solution

I really have no idea how to get started doing this problem. The first part of the problem asked "What is the potential difference V_cd?" This I found easy but once you close the switch I'm not sure how to approach the problem. Specifically how do you go about combining the capacitors to get a C_eq. Thanks for any help.

 

[tiny-tim]

Hi Black Armadillo!

Hint: the potential at c and d is the same, so draw the diagram again without the line cd, with the points c and d the same.