# Compute closed line integral of electric field in circuit

I have a circuit where resistor is parallel to capacitor, which is charged with voltage U. How to compute line integral around closed loop to get the result of Kirchhof second law - $U_{capacitor}+I\cdot R=0$?

For beginning I splited it in two parts, starting from the capacitor positive plate through resistor:
$$\oint \vec{E} \cdot d \vec{l} = \int_{+R-}\vec{E_{R}} \cdot d \vec{l} + \int_{-C+} \vec{E_C} \cdot d\vec{l}$$ Since intensity of resistor is with the same direction as current density $\vec{j}$: $$\int_{+R-}\vec{E_{R}} \cdot d \vec{l} = \int_{+R-} \rho \cdot \vec{j} \cdot d \vec{l} = I \cdot R$$ When I do the same for capacitor I gain negative line integral, because the $E_{c}$ is opposite to $d \vec{l}$ when positive charge is moved from negative plate to the positive.

So in which part I have made a mistake?

Assign potential $V=0$ to node "-", and potential $V_+$ to node "+". $V_+$ could be either positive or negative, but I'll be assuming positive for definiteness. Clearly, if you compute the line integral $\boldsymbol{E}$ around the circuit, you get a result of 0. I'll use the direction of integration in your question to analyze the elements of that integral. (You could take the opposite sense and get the same result.)
• Resistor. Your integral across the resistor, from node "+" to node "-", is correct, with result $IR$. If $V_+> 0$, the electric field will be parallel to $d \boldsymbol{l}$ and you find $I>0$, meaning current is flowing into the "+" node of the resistor, through the resistor, and out the "-" node.
• Capacitor, integrating from node "-" to node "+". I'll assign the capacitor potential $U_{capacitor} = V_+$, so that the capacitor's "plus" plate (which could have either sign of charge and voltage) is connected to node "+". Now, if $V_+=U_{capacitor}>0$ the electric field in the capacitor will point from the cap's "plus" plate to its "minus" plate, because $\boldsymbol{E} = - \boldsymbol{\nabla} U$ by definition (that is, $\boldsymbol{E}$ points from high $U$ to low $U$). Then, in your integral, the electric field in the capacitor will be anti-parallel to $d \boldsymbol{l}$, and you get a negative result for this contribution: $-U_{capacitor}. Adding the two pieces, which must sum to 0, you get: $$-U_{capacitor} + IR = 0 \text{ , or}\\ U_{capacitor} = IR$$ Finally, by convention one takes$q$to be the charge on the "plus" plate of the capacitor (here at node "+"), so that$q=C U_{capacitor}$. Then$dq/dt$represents the charge flowing into the capacitor's "+" node (through the part, and out the "-" node), which is the opposite of the resistor current: $$\frac{dq}{dt} = -I$$ Since$q= C U_{capacitor}$, the above loop equation can be re-written as: $$\frac{q}{C} = - \frac{dq}{dt} R \text{ , or} \\ \frac{q}{RC} + \frac{dq}{dt} = 0$$ Choose the negative plate of the capacitor as ground,$V=0$. The usual convention is that the electric field lines are drawn in the direction that a positive test charge would move if placed in the field. Then$\int_{-C+} d\vec{l}\cdot \vec{E}_C > 0$if I traverse the circuit from the negative plate to the positive plate of the capacitor. • If the capacitor with resistor would be charging, I would agree, but as capacitor is discharging the current is opposite. – Jānis Erdmanis Oct 24 '12 at 5:01 So after long I found why the equality: $$U_{capacitor} + I\cdot R =0$$ Is somehow correct for calculating passed charge and current in time when the differential equation is solved. The point is that the circulation is calculated correct in way: $$\oint_{R+C} \vec E \, d\vec l=\int_{+C-} \vec E_C\,d \vec l+\int_{-R+}\vec E_R\,d\vec l= [ \begin{array}f \vec E_C &\mbox{is with the same direction as d\vec l}\\ \vec E_R &\mbox{since \vec J = \sigma\,\vec E_R but current in chosen direction is opposite } \end{array} ] =\\ = \frac{q}{C} - I \cdot R$$ Which according to$\nabla \times \vec E =\vec{0}$is$0$. If the problem is to get charge$q$on capacitor at time t, the current is rewritten as$I = |\frac{dq}{dt}|$, but since q on capacitor decreases the$\frac{dq}{dt}<0\$ so the result: $$\frac{q}{C} + \frac{dq}{dt}=0$$ Which becomes my question if the current direction is chosen to charge the capacitor.