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Unformatted text preview: s the integral a bit easier to evaluate and
gives the same answer.) A)
B)
C)
D)
E) kq L / 2
dx L / 2 d L / 2 x
L
2kq L / 2 dx L 0 d x
kq L / 2 dx L L / 2 x d
kq L
dx 0 d L / 2 x
L
kq L dx L 0 x d V V
V
V V Version 1 Page 8 13. Charge is distributed on the surface of two very large insulating slabs such that
σ1=σ3=σ4=σ and σ2=σ. What is the electric field at point C?
To keep proper track of signs I follow this procedure:
1) draw Efield arrow for each sheet at point C
without considering the sign of the sheet (I treat
them all positive).
2) write an equation for E at point C. Arrows that
point left get a negative sign.
3) plug in the actual value of charge density for each
sheet (including the sign this time).
This way I count sign from charge only once. A) 0
B) E
2 0
C)
3
E
2 0
D)
3
E
2 0
E) E
2 0 Version 1 Page 9 14. Approximately what is the electric field at point P given that x>>d?
The x components cancel. We need to
find the y component. Due to
symmetry we can find y component of
field from posit...
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This test prep was uploaded on 03/22/2014 for the course PHYSICS 240 taught by Professor Davewinn during the Spring '08 term at University of Michigan.
 Spring '08
 DaveWinn
 Physics

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