- #1
mrlucky0
- 69
- 1
I am having a really tough time providing an argument for myself over why the answer is true:
The question:
A parallel plate capacitor is attached to a battery that supplies a constant potential difference. While the battery is still attached, the parallel plates are separated a little more.
Which statement describes what happens.
The solution:
ANSWER: Both the electric field and the charge on the plates decreases.
My reasoning:
The electric field decreasing, I can understand. I am imagining a small positive test charge inbetween the plates. As the plates are farther, the force on that test charge becomes weaker.
But what stumps me is why must the charges on the plates decrease? The problem states that the potential is kept constant, so what would influence the charges to decrease on the plates? If this can be reasoned using F=E*D, (F=force, E=electric field, D= distance) I'm certainly not understanding it.
The question:
A parallel plate capacitor is attached to a battery that supplies a constant potential difference. While the battery is still attached, the parallel plates are separated a little more.
Which statement describes what happens.
The solution:
ANSWER: Both the electric field and the charge on the plates decreases.
My reasoning:
The electric field decreasing, I can understand. I am imagining a small positive test charge inbetween the plates. As the plates are farther, the force on that test charge becomes weaker.
But what stumps me is why must the charges on the plates decrease? The problem states that the potential is kept constant, so what would influence the charges to decrease on the plates? If this can be reasoned using F=E*D, (F=force, E=electric field, D= distance) I'm certainly not understanding it.