Homework Help: Electric potential - plot,

The question is asking for the value of the electric potential at each point along the x-axis. You have three electric charges, so you are being asked to find the sum of the individual potentials of the charges. (The nice thing about doing this for electric potential is that you just have to add up numbers, rather than vectors.) So you don't want potential differences; you just want to calculate kq/r for each charge at any value of x and add up the values. (You will, of course, have infinities at the positions of the charges themselves...)

 

As I read your diagram, there are charges at x = 0, x = 4 cm, and x = 6 cm. The distances you use in your electric potential calculations are the distances of a chosen point x from each of these three positions.

The units for potential, BTW, are volts, since k has the units N·(m^2)/(C^2) = J·m/(C^2) = V·m/C .

If we want to find the total potential at, say, x = +1 cm, the distances from each charge will be 1 cm. = 0.01 m. from the charge at x = 0, 3 cm. = 0.03 m. from the charge at x = 4 cm., and 5 cm. = 0.05 m. from the charge at x = 6 cm. So the individual electric potentials at x = +0.01 m. will be

(9·10^9)·(2·10^-6)/0.01 m from the charge at x = 0 cm. ,
(9·10^9)·(-3·10^-6)/0.03 m from the charge at x = 4 cm. , and
(9·10^9)·(1·10^-6)/0.05 m from the charge at x = 6 cm.

You would then add up these values, one of which is negative, to find the total electric potential at x = +1 cm. You would do something similar for any other position x on the line. What you would plot would be the value of the total potential V (in volts) as a function of the position x. (You could also set up a set of terms for this and use it on a graphing program.)

At the position x = 0 , the distance from the charge at x = 0 is zero, so the potential would be (9·10^9)·(2·10^-6)/0 , which gives positive infinity. So your total potential will have "spikes" in it at x = 0, 4, and 6 cm., which each charge is located.

 

This is not calculating potential differences between points; we are calculating the value of total electric potential at a point x along the line which is not at the location of any of the charges. Each charge contributes a potential V = kq/r at a point x, given by

V_1 = k·q_1 / (x - 0) Volts,
V_2 = k·q_2 / (x - 0.04) Volts, and
V_3 = k·q_3 / (x - 0.06) Volts,

where x is the location of a point on the line, in meters.

The total electric potential is then V = V_1 + V_2 + V_3 . Since this sum depends on the value of x for a point we choose on the line, we say that V is a function of x.

What you are being asked to plot is that sum V against a position on the x-axis, represented by the coordinate x. The total electric potential will have positive and negative values (and be zero at some points), and will run away to positive or negative infinity at the location of the charges.

A potential difference would be the difference in electric potential between two locations on the x-axis. So, for instance, the potential difference between the point at x = -10 cm. and x = +10 cm. would be V(-0.1 m) - V(+0.1 m). You are not asked to plot that, but you could figure it out easily from your graph of V(x).

You are also not being asked to find the potential at the location of the charges: those would be positive or negative infinity.

 

The value of the potential is different for each value of x. Given x, it is possible to calculate the potential at x. We describe such a situation with the words: "the potential is a function of x".

They just want a graph with potential on the y-axis, and x on the x-axis. You've been shown how to calculate one point on the graph, the potential at x = 1 cm.

 

You do not put the value of x for each charge into the terms for the potentials; rather, you choose a particular position on the
x-axis and use its coordinate value to calculate each of the potentials. For x = + 1 cm = +0.01 m., this will give

V_1 = (9·10^9)·(2·10^-6)/|0.01 - 0| = +1,800,000 V ,
V_2 = (9·10^9)·(-3·10^-6)/|0.01 - 0.04| = -900,000 V , and
V_3 = (9·10^9)·(1·10^-6)/|0.01 - 0.06| = +180,000 V .

This makes the total electric potential at x = +0.01 m. equal to the sum,
V(+0.01 m.) = +1,080,000 V . (And I realized that the general expression should have absolute value signs, since we want the distance of the point x from each charge with no regard to sign.) You would do a similar calculation for every value of x.

As for your examples, you don't want to use x = 0, 0.04 m, or
0.06 m. The first of these will give you

for x = 0 m., V_1 = k·q_1 / (0 - 0) Volts = k·q_1/0 = +infinity!

By the same token, at x = 0.04 m., you'd have

V_2 = k·q_2 / (0.04 - 0.04) Volts = k·q_2/0 = -infinity ,
since q_2 < 0 .

This makes the total electric potential undefined at the locations of the charges, so we do not do the calculation at those places.

The point is that you use a single value for x in all three charge potential terms and add those values together to get a total, not separate values for each charge.

 

Yes, just so!

Keep in mind that the denominators of the potential terms use the absolute values of distance, because it doesn't matter for electric potential whether the location of the "test point" is to the left or right of the charge. You can even use something like Excel to calculate the values, since the algorithm is well-defined and you need to find many results to make a plot. If you use something which cannot give absolute values, there is a trick to "fool" the software (I had to use it on the graphing program I have). You write each potential term as

k · q / [sqrt{ (x - 0.04)^2 }] , for instance,

the idea being that the software always finds the positive square root and squaring the distance first eliminates the sign of the (x - 0.04) term, so the square root gives the absolute value |x - 0.04| .

You will find, I believe, that your potential function is zero at three positions. You may also want to plot the potential V in millions of volts (divide your calculated results by 10^6), because the values can get quite large (indeed, they will race off to infinity when you're close to any of the charges).