what q a a area of plate second integrate e to find

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: charged conductors: +Q y d σ E= εo E x -Q What is σ ?? What ?? Q σ= A A = area of plate Second, integrate E to find the potential difference V d d 0 V = − ∫ E ⋅ dy d 0 V = − ∫ (− Edy) = E ∫ dy = 0 Q d εo A As promised, V is proportional to Q !! 12 Q Q C≡ = V Qd / ε o A C= ε0 A d C determined by determined geometry !! geometry Physics 212 Lecture 7, Slide 12 Physics Question Related to Preflight +Q0 Initial charge on capacitor = Q0 BB d - Q0 +Q1 Insert uncharged conductor Charge on capacitor now = Q1 d t - Q1 How is Q1 related to Q0 ?? A. Q1 < Q0 B. Q1 = Q0 C. Q1 > Q0 14 Plates not connected to anything CHARGE CANNOT CHANGE !! Physics 212 Lecture 7, Slide 13 Physics Where to Start?? +Q0 d t BB - Q0 What is the total charge induced on the bottom surface of the conductor? A. B. C. D. E. 17 +Q0 -Q0 0 Positive but the magnitude unknown Negative but the magnitude unknown Physics 212 Lecture 7, Slide 14 Physics WHY ?? +Q0 E - Q0 E=0 +Q0 E - Q0 WHAT DO WE KNOW ??? E must be = 0 in conductor !! Charges inside conductor move to cancel E field from top & bottom plates 19 Physics 212 Lecture 7, Slide 15 Physics Calculate V y Now calculate V as a function o...
View Full Document

This note was uploaded on 02/09/2012 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.

Ask a homework question - tutors are online