An infinite sheet of charge, oriented perpendicular to the x-axis, passes throug
ID: 1273588 • Letter: A
Question
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density ?1 = -2.7 ?C/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.3 cm and b = 4.8 cm. The conducting slab has a net charge per unit area of ?2 = 86 ?C/m2.
1)What is Ex(P), the value of the x-component of the electric field at point P, located a distance 7.2 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 7.2 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 ?b, the charge per unit area on thesurface of the slab located at x = 4.8 cm?
6)What is Ex, the value of the x-component of the electric field at a point on the x-axis located at x = 3.3 cm ?
7)What is ?a, the charge per unit area on the surface of the slab located at x = 2.3 cm?
8)Where along the x-axis is the magnitude of the electric field equal to zero?
a.x < 0
b.0 < x < 2.3 cm
c.x > 4.8 cm
d.none of these regions
Explanation / Answer
Relevant equations
Gauss' Law E = ?/2*epsilon
3. The attempt at a solution
PART A
-- For x = -1 cm, I summed the two individual electric fields. For the sheet I said the electric field would be positive since the sheet's charge is negative so the field lines are going in the positive x direction (towards the sheet). For the slab I said the electric field was negative since the slab had positive charge, so the field lines are going away from the slab (i.e. in the negative x direction)
-- For x = 1 cm, same logic. I summed both individual fields again. This time both were negative since field lines for both the sheet and the slab were headed in the negative x direction.
-- For x = 2.5 cm, the electric field is zero since it is within the conductor and conductors have zero electric fields within them.
-- For x = 6 cm, I only used the electric field of the slab, since I thought it would block the field of the sheet. However I said it had a sigma of 2