Electric Field & Potential: Solve for Radius

[twisted079]

Homework Statement

The electric potential immediately outside a charged conducting sphere is 250 V, and 10.0 cm farther from the center the magnitude of the electric field is 440 V/m.
(c) Determine all possible values for the radius of the sphere. (Enter your answers from smallest to largest. If only one value exists, enter "NONE" in the second answer blank.)

Homework Equations

V = keQ / R and E = keQ / R²

The Attempt at a Solution

Im assuming that since there are two answers (Im sure it is two definite answers), it has to do with the squared distance in the denominator of E; (+/-) a certain number. There is most definitely proportionality here.

My attempt hasnt gotten me far. I found the electric potential at the +10cm where the electric field is 440. That hasnt gotten me anywhere. I try to set up a system of two equations where 1) 250 = keQ/R₁ ----> R = 210/keQ and 2) 440 = keQ/(R₂²+10) *radius should be given in cm*-----> (R₂²+10) = keQ/440

I have no idea where to go from here. My head hurts from this problem and I put so much time into it and got basically no where. Please help.

 

[ehild]

Homework Equations

V = keQ / R and E = keQ / R²

The Attempt at a Solution

I try to set up a system of two equations where 1) 250 = keQ/R₁ ----> R = 210/keQ and 2) 440 = keQ/(R₂²+10)

You found the correct equations for V and E, but you applied the second one wrong. In the first case, R₁=R, the radius of the sphere, but in the second case R₂=R+0.1. You have to use meters as the formula is valid for SI units. And you have to square R₂ which is R₂²=(R+0.1)².

So you have the equation

[tex]\frac{kQ}{R}= 250[/tex]

[tex]\frac{kQ}{(R+0.1)^2}= 440[/tex]

Eliminate kQ. You get a quadratic equation for R. Solve.

ehild