Potential difference between two assymetrically charged plates

3,021

Let us say plate $A$ has a charge $q_1$ and plate $B$, which faces plate $A$ has a charge $q_2$. By making use of the fact that the net field in the bulk of a conductor in static conditions is zero, and that the net field near the outer surface of a conductor equals $\sigma\over\epsilon_0$ [$\sigma=$local surface charge density], you can prove the following:

  1. Charge on the outer surfaces of $A$ and $B$ are each $q_1+q_2\over2$.
  2. By conservation of charge, the charges on the facing surfaces of $A$ and $B$ are $q_1-q_2\over 2$ and $-(q_1-q_2)\over2$ respectively.
  3. Now using the result for net field near the outer surface of a conductor mentioned above, the field in the region between the plates is, $\left|\overrightarrow E\right|= E ={{q_1-q_2\over2}\over A\epsilon_0}$, pointing towards plate $B$ if $q_1>q_2$. Here $A$ is the area of each plate.

Finally, potential difference is $E\cdot d$, where $d$ is the distance between the plates. This is valid because for infinitely large plates, the field in the region between the plates can be considered uniform. Since capacitance $C={A\epsilon_0\over d}$, we obtain the potential difference $V= {q_1-q_2\over 2C}$

Share:
3,021

Related videos on Youtube

abhijeetviswa
Author by

abhijeetviswa

Love to program.

Updated on December 26, 2020

Comments

  • abhijeetviswa
    abhijeetviswa almost 3 years

    What is the potential diff between two plates of charges $q_1$ and $q_2$ ($q_1$ $>$ $q_2$)? I know the equation is $V = \frac{q_1-q_2}{2C}$ where $C$ is the capacitance of the system but how has this equation been derived?

    I apologize if this is a duplicate question. Google searches with different search terms didn't yield any answers.