An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform
ID: 1327971 • Letter: A
Question
An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity 1 = 0.26 C/m2. Another infinite sheet of charge with uniform charge density 2 = -0.41 C/m2 is located at x = c = 37 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 16.5 cm and x = 20.5 cm).
1)
What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (8.25 cm, 0)?N/C
Answer: 3.78*10^4
2)
What is a, the charge density on the surface of the conducting slab at x = 16.5 cm? C/m2
Answer: -.335
3)
What is V(R) - V(P), the potentital difference between point P and point R, located at (x,y) = (8.25 cm, -20.5 cm)?
ANSWER: ? V
4)
What is V(S) - V(P), the potentital difference between point P and point S, located at (x,y) = (28.75 cm, -20.5 cm)?
ANSWER: ? V
5)
What is Ex(T), the x-component of the electric field at point T, located at (x,y) = (45.25 cm, -20.5 cm)?
ANSWER: ? V
HELP WITH 3, 4, &5 PLEASE
Explanation / Answer
(3)
the potentila difference across V(R) - V(P), is zero becuase the E field is only in the X direction and the path is all in the Y, integrating along it gives a potential change of zero
(4)
the potential difference between point P and point S
Break the path into two parts: P to R and R to S. As we saw in (3) the first part has no effect. The second will just be
V = -integra from R to s E dl = E l = 3.78*10^4(16.5 cm) = 6.23 * 10 ^3 V
(5)
the electric field of slabs is independent of distance
E = 1 / 2 episolon _0 -2 / 2 episolon _0
=0.26 / 2( 8.85 * 10 ^-12) - 0.41/2( 8.85 * 10 ^-12)
= -8.47 * 10 ^3 N/C