# So frustrated been at it for over an hour. electric field and electric potential HELP

• 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.

## 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

$$\frac{kQ}{R}= 250$$

$$\frac{kQ}{(R+0.1)^2}= 440$$

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

ehild