Capacitors - m having problem calculating the potential difference

[exuberant.me]

Q) In the adjacent circuit, find the potential difference across AB

Fine , now i did the charge distribution as follows

Now in the first loop (sorry i forgot to mention the points in my image)
applying kirchhoffs loop law
10 = q1/8 + q/8

or, q + q1 = 80

Next in the second loop

q1/8 + (q - q1)/8 + (q - q1)/8 = 0

or, 2q - q1 = 0
or, q1 = 2q

gives, q = 80/3 uc

Again in the second loop
V_A - (q - q1)/8 - q1/8 = V_B

=> V_A - V_B = q/8 = 10/3 V
but answer says its 2V.

Definitely i m wrong but where?

Any help is appreciated..!

 

[NaOH]

I'm not sure I follow the q-q1 arguments... and kirchhoff's loop rule seems to be producing more variables then equations for me, so I would suggest something else.

1) Ignoring AB for the moment, can you reduce the circuit into a simpler circuit with equivalent capacitance? Clearly draw out the transformation, because you'll need it to find Vab later!

The following rules might be useful:

$1/C_{1} + 1/C_{2} = 1/C_{eq}$
$C_{1} + C_{2} = C_{eq}$

But I'm leaving it to you to know when to use which!

 

[exuberant.me]

i don't knw why i always get those suggestions i probably knw.. :P

dear what i want you to do is check what's wrong in my soln...
if u cud please point out my mistake, that would help me clear 100 doubts of mine...
i obviously have an alternate soln.. the one you posted using it...
but they don't actually work in complex capacitor problems.
Thank you!

 

[gneill]

Check your equation for the second loop. It looks like you're going counterclockwise around the loop. Be sure to use the "leading" charge expression (expression for the charge on the first plate encountered on your KVL walk) for each capacitor. Could be you used the wrong one for the first term in your equation...

 

[exuberant.me]

i think i m pretty correct in my soln..
Rather the answer given is wrong...!

 

[gneill]

i think i m pretty correct in my soln..
Rather the answer given is wrong...!

"Pretty correct" is not quite the same thing as "correct"

The answer should be 2V as advertised.

 

[exuberant.me]

yup u r not "pretty right" u r "right" this time and me to.. lol.. :P

got it finally...

q + q1 = 80 and 2q = 3q1 and simply, Va - Vb = (q - q1)/8 = 16/8 = 2V.

Thanks!