Question

The plates of a capacitor are charged using a battery, and they produce an electric field across the separation distance d between them. The two plates are now to be pushed together to a separation of d/2. The pushing together can be done either with the battery connected or with it disconnected. In which case, with the battery connected or disconnected. Is the electric field magnitude greater with the battery connected with the battery disconnected?

Answer 1

Step-by-step explanation:

The equation of the capacitance of the capacitor can be represented as:

`C = (\varepsilon_oA)/(d)`

Also, the electric field between the plates can be expressed as:

`E = (\sigma)/(\varepsilon_o)`

`= (Q)/(A \varepsilon _o) \ \ (surface \ charge\ density \ \sigma =(Q)/(\varepsilon_o))`

However;

when pushed to a distance d/2, the new capacitance of the capacitor is:

`C = (\varepsilon _oA)/((d/2)))`

`=(2 \varepsilon_oA)/(d)`

`=2C \\ \\ Q' = CV \\ \\ = 2CV \\ \\ =2Q`

SImilarly, the new electric field between the plates is:

`E' = (\sigma')/(\varepsilon_o) \\ \\ = (Q')/(A \varepsilon_o) \\ \\ = (2Q)/(A \varepsilon_o) \\ \\ =2E`

For Battery disconnected:

The electric field between the plates doesn't rely upon the distance between the plates yet relies upon the magnitude of the charge. At the point when the battery is detached, the charge on the capacitor stays as before, so does the electric field.

Therefore, the magnitude of the electric field when the battery is associated is twice however much the magnitude of the electric field when the battery is separated and disconnected.

Hence, the ratio is :

`(E_(connected))/(E_(disconnected)) =(2E)/(E) \\ \\ (E_(connected))/(E_(disconnected)) = 2`

Hence, the ratio is = 2