Let's say two long cylindrical conductors of resistivity ρ1 and ρ2 are joined together and a current I is flowing uniformly through the cross section

enter image description here

So if we consider this scenario two different fields would be generated in the conductors due to the difference in their resistivity.


V1=Iρ1l/A V2=Iρ2l/A

E=V/l for both

further using gauss law at the cross section we can conclude a charge of magnitude ε0I[ρ1-ρ2] is accumulated at the cross sectional area.

My question is that how is the charge generated? Shouldn't charge conservation be considered as well if the current flowing in and out is the same ?

Pls provide an explanation along with experimental data if you have.

  • 1
    $\begingroup$ Charge need not be "generated", it is already there inside the conductor and on its surface, so part of it can concentrate also on the boundary between two different conductors. It does not contradict charge conservation in any way. $\endgroup$ Mar 17 at 14:15
  • $\begingroup$ @JánLalinský got that charge is not generated, but When opposite charges are separated, they create their own electric field, which can influence the accumulation of charge and affect the flow of current, right? $\endgroup$
    – Shravani
    Mar 17 at 18:14
  • $\begingroup$ Right, opposite charges that are not at the same position have electric field in their vicinity and this can act on any other charges in that region. $\endgroup$ Mar 17 at 22:24

1 Answer 1


You can think about it dynamically. I’ll restrict to 1D, so $\rho_1,\rho_2$ are linear resistivities (unit $\Omega/m$ in SI). Your two resistances have respective lengths $L_1+L_2=L$, resistances $R_1=L_1\rho_1,R_1=L_1\rho_1$ with currents $I_1,I_2$ and voltages $V_1+V_2=V$.

Say you impose a voltage $V$ and start with no charges accumulated at the junction. This means that the electric field is the same so: $$ E=\frac{V_1}{L_1}=\frac{V_2}{L_2} $$ Which gives: $$ \begin{align} V_1&=\frac{L_1}{L}V & V_2&=\frac{L_2}{L}V \\ I_1&=\frac{V}{L\rho_1} & I_2&=\frac{V}{L\rho_2} \end{align} $$ In particular $I_1\neq I_2$. To preserve conservation of charge, you deduce that you have an accumulation of charge $Q$ at the junction: $$ \dot Q=I_1-I_2 $$ Accounting in general for the excess charge, using Gauss’s law and assuming you are in the quasi static regime, the original equation becomes: $$ -\frac{V_1}{L_1}+\frac{V_2}{L_2}=\frac{Q}{\epsilon_0} $$ which gives: $$ \begin{align} V_1 &= \frac{L_1}{L}V-\frac{L_1L_2}{2\epsilon_0L}Q & V_2 &= \frac{L_2}{L}V+\frac{L_1L_2}{2\epsilon_0L}Q \\ I_1 &= \frac{V}{L\rho_1}-\frac{L_2Q}{2\epsilon_0\rho_1L} & I_2 &= \frac{V}{L\rho_2}+\frac{L_1Q}{2\epsilon_0\rho_2 L} \end{align} $$ This results in the ODE: $$ \dot Q +\left(\frac{L_2}{\rho_1}+\frac{L_1}{\rho_2}\right)\frac{Q}{2\epsilon_0L}=\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right)\frac{V}{L} $$

You therefore obtain and exponential relaxation to the equilibrium value: $$ Q_\infty=2\epsilon_0\frac{\rho_2-\rho_1}{R_1+R_2}V $$ Intuitively, in the stationary regime, the excess charge enhances the field in the region of high resistivity. Conversely, it suppresses the field in low resistance. This balances out the resulting currents stopping the charge pileup.

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.