Electric potential and work

1. Jun 7, 2008

scholio

1. The problem statement, all variables and given/known data

points a, b, and c have the following electric potentials:
V_a = -50volts, V_b = +2volts, V_c = -70volts

compute the work required by an external agen to transport the following charges at constant speed between the endpoints indicated:

a) +2coulomb from a to c
b) -2coulomb from c to a
c) -2coulomb from b to c

2. Relevant equations

acceleration a = (q/m)E = v^2/r where q is charge, m is mass, E is electric field , v is velocity in m/s ---> since the question asks for constant speed, acceleration = 0

electric potential difference V = kq[(1/r_a) - (1/r_b)] where q is charge, k is constant
9*10^9, r is radius/distance in meters

electric potential V = kq/r

work W = F*d, where F is force in newtons, d is displacement in meters

F = kq_1q_2/r^2 where k is constant, q is charge, r is radius/distance in meters

3. The attempt at a solution

my main issue is determining what equation to use, i am not sure about how to derive the correct equation.

it seems that i could use the electric potential difference equation because it involves one charge but i do not know the radii for the two potentials, can i determine the radii without a charge for each one?

i also need to calculate force, but the equations involved either involve wo charges, where i only am given one, or a radius, again which i am not given.

even still how do i calculate work, W = F*d

any tips on how to get started appreciated...

2. Jun 7, 2008

Dick

A volt is 1 Joule/coulomb. I.e. it takes one Joule to transport one coulomb over a potential difference of one volt. Forget about forces, fields, velocities and accelerations. Enuf said?

3. Jun 7, 2008

scholio

still not to sure on what to do, this is what i tried for part a) --? +2coulomb charge from A to C:

delta V_AC = V_C - V_A = kq/r^2
-70 - (-50) = (9*10^9)(2)/r^2
r^2 = [(9*10^9)(2)]/-20 = -9*10^8
r = 30000i where i means imaginary number

firstly i don't know how radius relates to work, nor how i got an imaginary number for the r value

could you please explain my errors...

4. Jun 7, 2008

alphysicist

Hi scholio,

The formulas:

$$E = \frac{k |q|}{r^2} \mbox{ and } V = \frac{k q}{r}$$
do not apply to this problem. Those are the fields and potentials produced by a point charge (and we can use them for field and potentials from a sphere of charge). Unless we know that the test charge is being affected by point charges, we would not use these.

The way to find the potential difference from a to c is much easier. What's the potential at c (the final point)? What's the potential at a? What's the difference in those two numbers?

Once you have that, how are potential difference and electric potential energy related? Then how is the work done by the external agent related to the change in energy?

5. Jun 7, 2008

scholio

sorry for double-posting, i was too late to edit the original post

actually i incorrectly entered the equation. it is is ' -kq/r^2 ' as opposed to ' kq/r^2 ' which i had originally, the correction makes r = +30000

though it still does not really help me, because i don't know how radius relates to work, other than it could be the displacement d in work W= F*d, but again Dick instructed me to disregard solving for force

so...

6. Jun 7, 2008

alphysicist

scholio,

I think if you answer the questions in my post you will understand why you cannot take that route in solving this problem. If something did not make sense let me know.

7. Jun 7, 2008

scholio

oh i think i get it now, i found an equation that i skipped over relating potential energy and electric potential difference

potential energy U_AC = U_C -U_A = -W_AC where U is potential energy, W is work

electric potential difference V_AC = U_AC/q

so i found the potential difference between A and C to be -70 - (-50) = -20volts

i subbed that into V_AC in the second eq, subbed in +2coulombs for q and solved for U_AC to get:

q(V_AC) = U_AC
+2(-20) = U_AC = W_AC
so work W_AC = -40 volts per coulomb or are my units coulomb-volts?

is that correct now, it seems right?

and for part b i got work W= -40 coulomb-volts, part c -->W = +144coulomb-volts

Last edited: Jun 7, 2008
8. Jun 7, 2008

alphysicist

Those numbers sound right. You could say the units of work are coulomb-volts; however, a volt is a joule per coulomb, so you can simplify that.

9. Jun 7, 2008

Dick

Yes, a coulomb*volt is a joule. I told you this a long time ago, but I'm glad you finally got it.