Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B . Then the capacitance of the system is

 Applying Gauss’s law to this closed surface, we get,


⇒ EΔA =σΔA/ε0

⇒ E = σ/ε0 = Q/Aε0

Now, the potential difference between the plates will be V=V+−V−


As you go from 2→1, the electric field and displacement are in opposite direction and therefore, E.dr = −Edr

V=−∫E.dr = ∫Edr = Ed =Qd/Aε0

Because capacitance is equal to QV, C=ε0A/d. Similarly, the capacitance of the other capacitor is C=ε0A/d. Because the positive and negative plates of both capacitors have similar points, the capacitances are in parallel. As a result, these two capacitors’ equivalent capacitance will be

C = C1 + C2 = ε0A/d + ε0A/d =2ε0A/d.Hence, the capacitance of the system is 2ε0Ad.

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.