Lab 202 - Part1

Lab 202 - Part1 - Ryan Johnson 2 3/12/2010.1 1 ε0 8 0 . 5...

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Unformatted text preview: Ryan Johnson 2 3/12/2010.1 1 ε0 8 0 . 5 Lab 202: Numerical Verification of Gauss Law 9 Part I (Charge inside cube): q 9.(1 ) 1 0 1 i a1 0 0 1 .ji ej f.k .k Pd 0 0 0 0 1 1 . 4. . π ε q 1 d 0e 0 f 0 123 ... .i y z rx,z p,y) P .j .k (,y) (x,z p,y) x ( ,z x E( x, , ) yz 0 ( r( x, , ) ) yz 3 .( x, , ) r yz ΦΦ Φ Φ ΦΦ Φ ul r l f b poi e r a wg f o c htnk t t a a Φ u p a a a a Φ ε0 r f o n t a 8 .1 .0 5 a 2 1 a a E( x,y , a ) . kd xd y Φ l o w a a E( a ,y ,z) . d y d z iΦ b a c k a a a E( x, a ,z) . jd xΦ dz r g i h t Φ e l t f a a E( x,y ,a ) .kd xd y a a a a a E( x,a ,z) . d xd z j a E( a ,y ,z) . id y d z a 9 q 9.(1 ) 3 1 0. Φ=1 1 . 00 7 Φ=2 . Φ =1 Φ Φ1=1 u.=2 .l 2 Φ =. 3 p 0 o . fb e . 3 w4 ra l8 9 9 3 o t0 6 . 5 f9 8 Φ di4ej .k = . n 4 i9 r 1 Φ 9 1tc gε h t k P. f 0 The total flux is the sum of the flux acting on an object on all sides 9 since flux = charge / epsilon, 1 q 1 . .( x, , ) E0, , ) (x y z r yz 4. . 0 ( r( x, in)the same magnitude as the charge that we started with π ε result , ) 3 epsilon * charge should yz a1 Part II (Charge outside cube): d 0e 0 12 .. f1 0 p,y) x y z rx,z p,y) P ( ,z .i .j .k (,y) (x,z x 1 i 0 0 0 j 1 0 0 k 0 1 Ryan Johnson 3/12/2010 Since point charge is located outside of the cube, there is no flux because the net charge inside of the cube is just 0. Conclusion Gauss’s law is independent of the shape or size of the object that is encapsulating the charge, all that it takes into account is the net charge inside the object. Since it only needs to account for the net charge inside, it does not matter where inside the cube the charge is located, the net flux will be the same. Questions: 2. If you change the sign of the charge, the sign of the flux will also change because the angle between the electric field lines and Surface Area Vectors will be approximately 180 degrees instead of 0, so the cosine will have the opposite sign therefore giving the dot product a sign change. Φ =3 p9 u. 4 Φ =0 ε0.Φ = 0 Ryan Johnson 3/12/2010 3. If you double the magnitude of the charge, the flux will double because according to Gauss’s Law flux = Q/ε, it implies that 2*Q/ε = 2*φ 4. Increasing the size of the box from a=1 to a=2 then a=10 finally a=100 should have no affect on the flux because the number of electric field lines hitting each face will still cancel each other out ...
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This note was uploaded on 09/16/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.

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