Solving Conductor Questions: Surface Charge, Electron Count, Electric Field

[lha08]

Homework Statement

The potential of a metal sphere of diameter 2 cm is 10^4 V relative to the ground.
a) What is the surface charge density?
b) How many electrons were removed from the sphere?
c) What is the electric field strength at the surface?

Homework Equations

The Attempt at a Solution

a) so first I converted the diameter into the radius (0.01 m). Then I considered the formula
Q (total charge)= 4 X pi X R^2 X sigma. The sigma is the surface charge density. So then since i have the electric potential and the radius, I replaced Q in the first equation with the electric potential. In this case, VR/k =Q where k=9.0X10^9...and when I solved it, I got 8.84X10^-6 C/m^2...but does this seem right because I have a gut feeling that it isn't...

b) for the next question, i solved for Q (the total charge) and then converted the Joules into electrons by dividing the total charge by 1.602X10^-19 C...again I'm not sure if my method makes sense...

c) for the last one, I used E= kq/r^2..and I basically plugged in the numbers but I'm not sure if placing 0.01 m for the radius is the right move...

Any help would be appreciated!

 

[ideasrule]

a) so first I converted the diameter into the radius (0.01 m). Then I considered the formula
Q (total charge)= 4 X pi X R^2 X sigma. The sigma is the surface charge density. So then since i have the electric potential and the radius, I replaced Q in the first equation with the electric potential. In this case, VR/k =Q where k=9.0X10^9...and when I solved it, I got 8.84X10^-6 C/m^2...but does this seem right because I have a gut feeling that it isn't...

It's right. The static charge that builds up on everyday objects is extremely small, on the order of millionths or billionths of a coloumb. Even a cloud-to-ground lightning strike delivers only 300 coloumbs.

b) for the next question, i solved for Q (the total charge) and then converted the Joules into electrons by dividing the total charge by 1.602X10^-19 C...again I'm not sure if my method makes sense...

Well, you didn't convert the joules to electrons; you converted the coloumbs to electrons. Your method is right.

c) for the last one, I used E= kq/r^2..and I basically plugged in the numbers but I'm not sure if placing 0.01 m for the radius is the right move...

Yup, that's how you do it.