Parallel plane - electric field and potential

[Deathnote777]

parallel plane -- electric field and potential

Homework Statement

Consider 2 plate A and B. They are connected with a battery and their distance is halved. What will happen as a whole ?

Homework Equations

The Attempt at a Solution

1. Charge per unit area increase due to increase in electric field.
2. I think the potential difference will not change although there is increase in charges because both A and B 's potential has increased by the same amount. So there will be no change in potential difference.
Am I right ?
I have another question.
1. If one of the plates(connected with battery) is earthed and they are put closer, what would happen ? Will the potential difference, number of charges and electric field change ?
2. If the amount of charges in the plates are fixed. One of the plates,A, is earthed originally. When B is approaching A, the number of charges in A is still constant, but why the potential of B will increase? Isn't it that only the number of charges determines a plate's potential ?Thank you

 

[haruspex]

1. Charge per unit area increase due to increase in electric field.
2. I think the potential difference will not change although there is increase in charges because both A and B 's potential has increased by the same amount. So there will be no change in potential difference.

Right answer, wrong reason. One has a positive potential and the other negative. If they both increase, keeping a constant difference, then the positive will be more positive and the negative less negative. That fails on grounds of symmetry.
Neither potential changes, yet the charges increase. Do you see how that can be?

1. If one of the plates(connected with battery) is earthed and they are put closer, what would happen ? Will the potential difference, number of charges and electric field change ?

The P.D. is still controlled by the battery, so won't change. As before, the charge will increase.

2. If the amount of charges in the plates are fixed. One of the plates,A, is earthed originally. When B is approaching A, the number of charges in A is still constant, but why the potential of B will increase? Isn't it that only the number of charges determines a plate's potential ?

If A is earthed, its charge will change. Did you mean the charge of B is constant? What do you mean by 'the potential of B will increase'? Do you mean assuming it was positive? What makes you think it would increase? I think the P.D. will decrease.

 

[Deathnote777]

Right answer, wrong reason. One has a positive potential and the other negative. If they both increase, keeping a constant difference, then the positive will be more positive and the negative less negative. That fails on grounds of symmetry.
Neither potential changes, yet the charges increase. Do you see how that can be?

I think I am confused here. Why the potential of a plate will not change when the charge density has increased ? Assume there is one plate only. To my knowledge, increase in charge density in the plate will increase the force exerted on a charged object next to the plate. Hence, potential of that plate is increased, isn't it ?

f A is earthed, its charge will change. Did you mean the charge of B is constant? What do you mean by 'the potential of B will increase'? Do you mean assuming it was positive? What makes you think it would increase? I think the P.D. will decrease.

I assume it is negative

 

[haruspex]

I think I am confused here. Why the potential of a plate will not change when the charge density has increased ? Assume there is one plate only. To my knowledge, increase in charge density in the plate will increase the force exerted on a charged object next to the plate. Hence, potential of that plate is increased, isn't it ?

Yes, but the potential at a point in space is a result of the potentials due to all charges, including the other plate. As the plates are brought closer, the potential due to the other plate is strengthened. This is how capacitors work - they store a high charge without producing such a large voltage.