Solving for potential using surface charge density of a sphere

[BuggyWungos]

Homework Statement

I'm trying to solve for the potential of a charged copper sphere with only radius and electric field strength known. The field lines are directed into the sphere.
Radius of the sphere: 0.2 m
Electric Field Strength at the surface of the sphere: 3800 N/C
Answer = half of my solution's value.

Relevant Equations

rho = Q/A
Surface area = 4pi(r^2)
Electric Field strength = rho/2(epsilon nought) OR kQ/r^2
Electric potential = kQ/r

surfafce area = 0.502

E = -q/A2(en) = 3800
-q = 3800*(A2(en))
-q = 1.68*10^(-8)
-q = 3.37*10^(-8)

V = kq/r
V = (9.0*10^9)(-3.37*10^(-8))/0.2
V = -1519 V

 

[TSny]

Electric Field strength = rho/2(epsilon nought)

This is not the correct formula for the electric field at the surface. This is where your mistake of a factor of 2 occurs.

 

[BuggyWungos]

This is not the correct formula for the electric field at the surface. This is where your mistake of a factor of 2 occurs.

What is the correct electric field strength formula using rho? I understand that E = rho/(epsilon nought) would give the correct answer, but the formula I was given in my textbook was E = 2(pi)k(rho), which would simplify to E = rho/2(epsilon nought). Is the above formula used for another situation?

 

[TSny]

What is the correct electric field strength formula using rho? I understand that E = rho/(epsilon nought) would give the correct answer, but the formula I was given in my textbook was E = 2(pi)k(rho), which would simplify to E = rho/2(epsilon nought). Is the above formula used for another situation?

$E= \dfrac{ \sigma}{2 \varepsilon_0}$ gives the field of an infinite plane with uniform surface charge density $\sigma$. (The symbol $\rho$ is more often used for a volume charge density rather than a surface charge density.)

The field at a point just outside the surface of a conductor in electrostatic equilibrium is $E=\dfrac{\sigma}{\varepsilon_0}$. This can be derived using Gauss’ law.