A parallel-plate capacitor with air between the plates is connected to a battery

Question

A parallel-plate capacitor with air between the plates is connected to a battery that has a voltage of Vi. The capacitor has an initial capacitance of Ci. The capacitor stores an initial charge Qi, has an initial electric field of magnitude of Ei, and stores an initial energy Ui. The following steps are then carried out, one after the other.
Step 1 - with the capacitor still connected to the battery, the distance between the plates is reduced by a factor of 2.
Step 2 - after completing step 1, the wires connecting the capacitor to the battery are removed. Then, the distance between the plates is returned to its original value.
Step 3 - after completing step 2, a piece of dielectric material is inserted into the capacitor, completely filling the space between the plates. The dielectric constant is 2.0.
Fill in the table below to show the potential difference, capacitance, charge, magnitude of the field inside the capacitor, and energy stored by the capacitor, in terms of their initial values, after each step. Note that you just have to fill in numbers in the boxes below. If, for instance, you think the potential difference after step 1 is 3Vi, you enter 3 in that box.

Explanation / Answer

Potential Difference Capacitance Charge Electric field Energy after step 1 1 Vi 2 Ci 2Qi 2 Ei 2 Ui after step 2 2Vi 1 Ci 2Qi 2 Ei 4 Ui after step 3 1Vi 2 Ci 2Qi 1 Ei 2 Ui That is the chart... I am doing the same w.a. If you want we can help each other. I have two left

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