Question

Please show me how to do question 4 and 5. I would really appreciate your help.

Answer 1

@UnkleRhaukus

Answer 2

q4- in electrostatics, the net charge inside a conductor is zero.since -q is there at the center,an opposite charge +q is induced on the inner surface of the sphere such that the net charge inside the sphere=0

Answer 3

oh okay, yes i thought the answer would be +q as well.... just thought the question is meant to be more complicated than that. Thanks @MotherOfGod It would be really helpful if you could try question 5. Im not quite sure how it should be solved.

Answer 4

@UnkleRhaukus yep i understand the diagram, thanks, and where do i go from there?

Answer 5

can you tell me the magnitude of electric field in the center of the sphere?

Answer 6

uhm, would it be p ? since the charge distribution is p Cm^-3

Answer 7

it should be zero at the center,since the sphere is a symmetrical object

Answer 8

construct a spherical gaussian surface of radius r<R

Answer 9

@MotherOfGod This is clearly marked as an assignment , please dont reveal too much

Answer 10

you know gauss's law right? @Tushara

Answer 11

yep i do

Answer 12

\[\oint\vec E\cdot\mathrm d A=\frac{Q_{enclosed}}{\varepsilon^{}_0}\]

Answer 13

I know how to use equations but I just dont know what would happen in a situation like that....

Answer 14

can you find the electric field if r>R ? its a bit easier

Answer 15

in this case, net charge enclosed is 0.... Q/e = 0

Answer 16

nope, its only zero if the Gaussian surface is a point

Answer 17

lets work out the E field out side the sphere first

Answer 18

so is it flux/area?

Answer 19

so start with this,\[\Phi_E=\oint\vec E\cdot\mathrm d A=\frac{Q_{\text{enclosed}}}{\varepsilon^{}_0}\]

we have to find \(\vec E\)

Answer 20

E = Q/4(pi)(e)(r^2)

Answer 21

i see... but how do i know what will happen to the charge whenits released?

Answer 22

well to know how its going to move you, need to know the force on it, and to know that , you will need to know the Electric field

Answer 23

F = Qq/4(pi)(e)(r^2)