Two thin plastic spherical shels (shown in cross section in the figure below) ar

Question

Two thin plastic spherical shels (shown in cross section in the figure below) are unformly charged. The center of the larger sphere is at (o, o); it has a radius af 12 cm and a uniform positive charge ot +2 * 10 sphere is a (27 em, O: it has a radius of 3 cm and a uniform negative charge of -2 x 10-9 C C. The center of the smaller (a) What are the components Ex and EAy of the electric field E at location A (6 cm to the right of the center of the lerge sphere)? Neglect the small contribution of the polarized molecules in the plastic, because the shells are very thin and don't contain much matter. (Indicate the direction with the sign of your answer.) N/C NIC The number of significant digits is set to 3, the tolerance as +/-2% e the center of the small sphere)? Neglect the small contribution of the poliarized molecules in the plastic, because the shells are very thin and don't contain much matter. (Indicate the direction with the sign of your answer) The number of signifiant dgts set to 3, the teierance is +/-2% NIC (e) What are the components F, and F, of the force on an electron placed at location B? (Indicate the direction with the sign of your ansner.)

Explanation / Answer

(a)

Electric field at A due to larger sphere = 0

Electric field at A due to smaller sphere,

Ex = kq / r^2

Ex = 9*10^9*2*10^(-9) / (0.25 - 0.06)^2

EAx = 408 N/C

EAy = 0

(b)

At point B,

Electric field due to large sphere,

EBx = (kq / r^2) * cos(theta)

where, r^2 = (0.27^2 + 0.15^2) = 0.0954 m

cos(theta) = 0.27 / (0.27^2 + 0.15^2) = 2.83

EBx = (9*10^9*2*10^(-9) / 0.0954) * 2.83

EBx = 533.9 N/C

EBy = (kq / r^2) * sin(theta)

EBy = (9*10^9*2*10^(-9) / 0.0954) *  (0.15 /  (0.27^2 + 0.15^2))

EBy = 296.66 N/C

Electric field due to smaller sphere,

EB' = kq / r^2 = 9*10^9*2*10^(-9) / (0.15)^2

EB' = 800 N/C

x componenet of elecric field at point B,

EBx = 533.9 N/C

y componenet of elecric field at point B,

EBy = 296.66 - 800 = 503.34 N/C

(c)

Fx = qEx = 1.6*10^(-19)*533.9

Fx = 8.54*10^(-17) N

Fy = 1.6*10^(-19) * 503.34

Fy = 8.05*10^(-17) N

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