This preview shows page 1. Sign up to view the full content.
45. (a) The electric field in the region between the plates is given by E = V / d , where V is the potential difference between the plates and d is the plate separation. The capacitance is given by C = κε 0 A / d , where A is the plate area and κ is the dielectric constant, so 0 / dA C = and E VC A == × ×× =× − −− 0 12 12 4 2 4 50 100 10 54 885 10 100 10 10 10 VF Fm m Vm b gc h c hc h .. (b) The free charge on the plates is q f = CV = (100 × 10 –12 F)(50 V) = 5.0 × 10 –9 C. (c) The electric field is produced by both the free and induced charge. Since the field of a large uniform layer of charge is q /2 ε 0 A , the field between the plates is E q A q A q A q A ff ii =+−− 2222 0000 εεεε , where the first term is due to the positive free charge on one plate, the second is due to
This is the end of the preview. Sign up to access the rest of the document.
This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
- Spring '10