***Please explain!! You don\'t have to plug in the numbers, just help me with wh
ID: 1279757 • Letter: #
Question
***Please explain!! You don't have to plug in the numbers, just help me with what needs to be done and why... :)***
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density ?1 = -3.6 ?C/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.3 cm and b = 4.2 cm. The conducting slab has a net charge per unit area of ?2 = 60 ?C/m2.
1)What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.6 cm from the infinite sheet of charge?
2)What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.6 cm from the infinite sheet of charge?
3)What is Ex(R), the value of the x-component of the electric field at point R, located a distance 1.15 cm from the infinite sheet of charge?
4)What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.15 cm from the infinite sheet of charge?
5)What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.15 cm from the infinite sheet of charge?
6)What is Ex, the value of the x-component of the electric field at a point on the x-axis located at x = 3.06 cm ?
7)What is ?a, the charge per unit area on the surface of the slab located at x = 2.3 cm?
Explanation / Answer
Ex(P) = (sigma1) / (2 epsilon0) + (sigma2)/(2 epsilon0)
= (-3.6 * 10-6) / (2 * 8.85 * 10-12) + (60 * 10-6) / (2 * 8.85 * 10-12)
= 3.19 * 106 N/C
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Ey(P) = 0 N/C
[in an infinite sheet, have component in x direction]
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Ex(R) = (sigma1)/(2 epsilon0) - (sigma2)/(2 epsilon0)
= (-3.6 * 10-6) / (2 * 8.85 * 10-12) - (60 * 10-6) / (2 * 8.85 * 10-12)
= -3.59 x 106 N/C
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Ey(R) = 0 N/C
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Ex(P) = (sigmab)/(epsilon0)
sigmab = epsilon0 * Ex(P) = 8.85 * 10-12 * 3.19 * 106
= 28.2 * 10-6 C/m2
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Ex = 0 N/C
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sigmaa = (60 * 10-6) - (28.2 * 10-6 )
= 31.8 x 10-6 C/m2
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