(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A is an earthed metal plate while B is a positively charged metal plate parallel to A. The separation between the plates is much less than the length of each plate. If A is moved closer to B, what would happen to the electric potential of B and the electric field strength between the plates?

2. Relevant equations

I haven't pick the quick capacitor equations as i think there are differences and the equations:

1.E= V/d, are not practical here as both V and d decreases.

2.V = Q/C, i doubt if Q change

At last, I use E= Q/(AK), where K is the permitivity

3. The attempt at a solution

1. As A which is negatively charged moves closer to B, it solely establish a more negative potential at B and the resultant of the potential decreases.

2.On the other hand, the voltage of A should increase as it is closer to B, but as A is earthed, there are some negative charges flow into A to make it 0V, i.e. unchanged.

Applying E = Q/(AK) we know the E field produced by A increases and that of B is constant as charge on B is constant.

i.e. E resultant increases

My teacher told me to treat it like a capacitor and as the capacitor is on a open circuit, there must not be changes on Q, i.e. (Q1-Q2)/2. As a result of V=Q/C decreases and E=Q/(AK) remains unchanged.

Here comes my actual questions:

1. Is there really no charge flow even when one plate is earthed?

2. what is wrong with my approach?

Note:

1.the model answer: V decreases, E constant.

2. My teacher also told me to think in another way, the earthed A plate is just to make the potential reference constant. and we can see both plates isolated, in this case V can be seem as the potential difference of the plates.

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# Homework Help: A capacitor-like system, is this really a cap?

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