A circuit has a section `Ab` shown in fig. The emf of the source equals `E = 10V`, the capacitances are equal to `C_(1) = 1.0 muF` and `C_(2) = 2.0 muF`, and the potential difference `varphi_(A) - varphi_(B) = 5.0 V`. Find the voltage across each capacitor.

Explanation

Let, us make the charge distribution, as shown in the figure.
Now, `varphi_(A) - varphi_(B) = (q)/(C_(1)) - xi + (q)/(C_(2))`
or, `q = ((varphi_(A) - varphi_(B)) + xi)/(C_(1) + C_(2)) C_(1) C_(2)`
Hence, voltage across the capacitor `C_(1)`
`= (q)/(C_(1)) = ((varphi_(A) - varphi_(B)) + xi)/(C_(1) + C_(2)) C_(2) = 10 V`
and voltage across the capacitor `C_(2)`
`(q)/(C_(2)) = ((varphi_(A) - varphi_(B)) + xi)/(C_(1) + C_(2)) C_(1) = 5 V`