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i Safari File Edit View History Bookmarks Window Help ? 6s lD us.ntemnona..PC Tue 10:22 AM Shoug Q. 10ncapa2.physics.sc.edu Elimination Calculator-Solve System of Equations with Circuit Law and Kirchhoffs Circuit Main Menu Course Contents Course Contents» » HW 3 P064.problem Notes Bookmark è Evaluate Communicate Print Info Guitar chords with lyrics guitartabsexplorer.com Learn to play "Moonlight 2"easy by Grace Vand... A parallel plate capacitor with a plate area of 10.0 cm and air in the space between the plates, which are separated by 2.5 mm, is connected to a 20.0-V battery. If the plates are pulled back so that the separation increases to 5.9 mm, how much work is done? Submit Answer Tries 0/99 Home eyintranet.etihad ae This discussion is closed Send Feedback Turn on more accessible mode Skip t Outlook Web App webmail etihad.ae Security ( show explanation) This is a... pdf.z docs.google.com all.pdt Open with https:/l shopandship.com Live Chat English About Features Help Login S.. 6

Explanation / Answer

capacitance C =A/d

= 8.85 x 10-12

A = area

d= seperation

energy stored in a capacitor E = 1 CV2

work done W = change in the energy of the capacitor as it has gone from configuration 1 to 2

= (1/2) [ C1V2 - C2V2] --------------1

now , V = 20 v [ voltage remain same for both the cases]

C1 = 8.85*10-12 x 10 *10-4 / (2.5 x 10-3) = 3.54 x 10-11 F [ all the unit have been changed to meter]

C2 = 8.85*10-12 x 10 *10-4 / (5.9 x 10-3) = 1.55 x 10-12 F

from equation 1 , putting the values

W = 0.5* 202[ 3.54 x 10-11 - 1.55 x 10-12 ] = 6.77 x 10-9 J = 6.77 nJ