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, a second capacitor, identical to the first but initially uncharged, is placed next to the first capacitor so they are touching. This effectively doubles the area of the capacitor plates. The capacitor goes from looking like one equals sign () to looking like two equals signs), but with the equals signs touching. In steps 2 and 3, the capacitor plates remain twice as large as they were before step 1 Step 2 - After completing step 1, the wires connecting the capacitor to the battery are removed; and then the distance between the plates is increased by a factor of 4 Step 3 - After completing step 2, the capacitor is re-connected to the battery. 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 Potential Difference Capacitance Charge Electric field Energy After Step 11 After Step 2 4 After Step 3 1 Uj E 8 Ui V8

Explanation / Answer

step 1 , Electric field
Electric field E = V/d
as there is no change in v and d, there is no change in E
Hence E = 1 Ei

step 2 , Capacitance
Capacitance = A eo/ d
After step 1, area is two times so C1 = 2Ci
After step 2 d becomes 4 times , so C2 = 2Ci/4 = Ci/2
so answer is 1/2

step 3 , Capacitance
No change in area of plates or distance between plates. So Capacitance does not change in step 3.
So answer is 1/2

Step 3 , charge
q = CV, before step 1
Qi = CiVi
At the end of step 3 , V is still Vi, but C is Ci/2
so Q = Qi/2

step 3 , electric field
Electric field E = V/d
V is same but d is four times
so E = Ei/4

step 3 energy
Energy U = CV2/2
C becomes Ci/2, so energy becomes Ui/2

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