Oil Capacitor
Parallel Plate Capacitor help?
A parallel plate capacitor in air has a plate separation of 1.5cm and a plate area of 25cm^2. The plates are charged to a potential difference of 250V and then disconnected from the source. The capacitor is then immersed in silicone oil. (k=2.5, dielectric constant is 15x10^6 V/m) Determine:
a. the charge on the plates before and after the immersion
b. the capacitance and potential difference after the immersion
c. the change in energy stored in the capacitor
Please help, I was absent when the teacher explained how to do this.
A. Charge remains constant.
B. Q = VC → V = Q/C. Capacitance increases by a factor or 2.5, so voltage decreases by the same factor to 0.4 * 250 V.
C = e0*k*area/separation. Initially k = 1, after addition of oil = 2.5. See ref. 1.
C. E = QV/2 → E2/E1 = V2/V1 = 0.4. Energy decreases to 0.4 * its original value. Where did the rest of it go? It actually is converted to kinetic energy which "pulls" the dielectric into the space between the plates. With a solid dielectric the force of this pull is measurable (ref. 2).
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