# Potential difference across a capacitor Look at the above circuit, I didn't get anything of it. Firstly i tried it using Kirchoff's voltage rule, I failed. I'm confused with those batteries. Can someone please explain me the role of those batteries and the solution to this problem.

• V1 may be $$\frac{(E2-E1)C2}{(C1+C2)}$$ – Nehal Samee Feb 24 '18 at 10:14
• may be? ........ – Selena Feb 24 '18 at 10:16
• I'm only confused with those batteries... – Selena Feb 24 '18 at 10:17
• The term Cx in the question is unclear ... So I found the individual potential difference of the capacitors ... – Nehal Samee Feb 24 '18 at 10:18
• $C_x$ means $C_1$ and $C_2$ – Selena Feb 24 '18 at 10:18

## 3 Answers

Define $$V_1$$ and $$V_2$$ such that if $$q>0$$, then $$V_1>0$$ and $$V_2>0$$ (i.e. $$V_1$$ is the left plate potential minus right plate potential, and $$V_2$$ is the right plate potential minus left plate potential). Write KVL, adding voltage drops along the circuit: $$-E_1 + V_1 + E_2 + V_2 = 0$$ Since the charge induced in the capacitors are equal, $$q=C_1 V_1 = C_2 V_2 \Rightarrow V_1 = \frac{C_2}{C_1}V_2, V_2=\frac{C_1}{C_2}V_1$$ To find $$V_1$$, substitute the expression for $$V_2$$ in the KVL equation. This yields $$V_1 = \frac{C_2}{C_1+C_2}(E_1 - E_2).$$ Similarly, substituting the expression for $$V_1$$ gives $$V_2 = \frac{C_1}{C_1+C_2}(E_1 - E_2).$$ See..this is the answer! U understood? if yes, please clear, me not.

• Well...It's the direct formula...What's new...? – Nehal Samee Feb 24 '18 at 10:33
• I only not got how (1) Came, everything else is alright...the last part – Selena Feb 24 '18 at 10:34
• They are trying to say that , $V_2$ is connected at positive plates of both cells and so have a positive potential...Opposite for $V_1$... – Nehal Samee Feb 24 '18 at 10:36
• i'm not understanding what you are even saying, can you please elaborate the concept and be specific! – Selena Feb 24 '18 at 10:42
• Are you not understanding the loop part ? – Nehal Samee Feb 24 '18 at 10:47 The charge polarity on the capacitor due to the cell $E_1$ is not a problem I think . Excess electrons flow from the negative plate and flow into the positive plate . But your confusion lies in the cell $E_2$ . As you notice that $E_2$ has a higher tendency to flow current in its own direction . From the figure , electricity flows from + of top capacitor to - of bottom capacitor . And this direction matches with the current flow for cell $E_2$ . I think the opposite polarity is justified ...