TZ Exam 2 - 1(1 Altering capacitors A parallel-plate capacitor whose capacitance is 120pF is charged by a battery to a potential difference of 6V

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 (1) Altering capacitors A parallel-plate capacitor whose capacitance is 120pF is charged by a battery to a potential difference of 6V between its plates. The battery is now disconnected, and a slab of a dielectric material, whose dielectric constant is 80, is slipped between the plates. What is the potential energy, in joules, of the capacitor-slab device after the slab is inserted? (1) 2.7×10-11 (2) 5.8×10-7 (3) 1.5×10-3 (4) 0 (5) 7.4×10 3 Solution When a parallel-plate capacitor is altered in any way, the first thing to do is compare its final capacitance to its initial capacitance using the parallel plate equation. Initial Final 1 i A A C d d κε ε = = 80 80 80 f i A A C C d d ε ε = = = Note that in calculating the initial capacitance, we assume the dielectric constant to be 1. This is always a valid assumption unless you’re told that the original capacitor contains a dielectric. So now we know that 80 f i C C = . After comparing the capacitances, we then need to compare the final and initial energies. There are three equations to choose from: 2 2 1 1 2 2 2 q U qV CV C = = = Of all the variables, we know what happens to C (it goes up by a factor of 80). Since the capacitor is disconnected from the battery in the second half of the problem, we know also that the voltage V must change. The charge q on the other hand, must remain constant, because the plates of the capacitor were not discharged. In short, we know the behavior of q and of C , so we choose the rightmost of the equations. Again, we can compare initial and final: Initial Final 2 2 i i q U C = ( 29 2 2 2 1 1 2 2 80 80 2 80 f i f i i q q q U U C C C = = = = Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com For evaluation only. 2 So the final energy is 1/80 the initial energy. Often, a fractional answer like this is what you need to find in this type of problem. However, we have to go one step further in this one. To calculate the initial energy, we have the initial voltage and initial capacitance. Thus, we find the initial energy with the equation: ( 29 ( 29 2 1 2 2 12 1 2 9 120 10 6 2.16 10 J i i i U CV U U-- = = × = × The final energy should be this 2.16×10 –9 J divided by 80. Thus, we get 2.7×10 –11 J. Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com For evaluation only. 3 (2) Altering capacitors A parallel-plate capacitor has plates of area A = 63cm 2 and separation d = 5mm is charged to a potential difference V = 40 Volts. The charging battery is then disconnected, and the plates are pulled apart until their separation is 2d. Find the work required to separate the plates (in joules)....
View Full Document

This note was uploaded on 11/03/2010 for the course PHY 2049 taught by Professor Any during the Fall '08 term at University of Florida.

Page1 / 42

TZ Exam 2 - 1(1 Altering capacitors A parallel-plate capacitor whose capacitance is 120pF is charged by a battery to a potential difference of 6V

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online