Static Electricity Legacy Problem #18 Guided Solution

There are many things which make physics problem solving difficult, and for many students one of the sources of difficulties has to do with algebraic manipulation of equations, simply doing the proper algebra in solving a problem. But many students don't recognize that the other thing that makes problems difficult is difficulties in conceptual understanding of the physics of a problem. Here the problem has to do with an electric field surrounding a source charge. When we speak of an electric field, what we mean is that a source of charge alters the space which surrounds it such that any other charged object brought into that space would experience a force of attraction or repulsion. And we can describe the electric field in the space surrounding an object as simply the force per charge, the F per Q, where the Q is simply the quantity of charge on the so-called test charge that's placed in that space. So as you begin to solve electric field problems, you need to distinguish between the so-called test charge and the so-called source charge, the source charge being the charge that creates the electric field, the test charge being the charge you put in the space surrounding it in order to test the strength of the electric field. And here what we have are both charges being mentioned in the same problem, and what we know is that there is a force upon the test charge. The force is 4.0 times 10 to the negative 5th, and the test charge is 2.18 times 10 to the negative 8th. So in Part A, if we wish to calculate the electric field, what we have to do is we have to take the ratio of F per Q. So if we divide that F by that Q, we'll be able to get the electric field, and for me it comes out to be 2064.2202, and not all of those digits are significant. I'm going to round it to three significant digits such that it becomes 2060, or 2.06 times 10 to the 3rd Newtons per Coulomb. That's the magnitude of the electric field. By definition, we describe the direction of the electric field as simply being the direction of force upon any positive test charge. Here our test charge is negative, and it's being pushed northward. So we would imagine that a positive test charge would be pulled southward, and so we describe the direction of the electric field at this location as southward because that's the direction the positive test charge would be pushed or pulled. In Part B of this problem, what we have to do is determine the magnitude and the type of charge on the source. So now we've got a couple of options. We could use our Coulomb's Law equation, or we could use our other equation derived from Coulomb's Law that describes the electric field strength in terms of the quantity of charge on the source. That equation can be found on the overview page in this set of problems, and it goes E, electric field strength, which we just calculated, is equal to k times big Q, the quantity of charge on the source, divided by the distance squared from the source charge. Here the distance squared is 25.0 centimeters, or preferably 0.250 meters, and the k is a constant value of 8.99 times 10 to the 9. So if I use my E value calculated from Part A, the unrounded E value, and I plug it into the equation, I make an effort to solve for big Q, the quantity of charge on the source, I would end up getting 1.4351 blah, blah, blah, times 10 to the negative 8th Coulomb's. I can round that to two significant digits, such that it becomes 1.44 times 10 to the negative 8th Coulomb's. Finally in Part C, what I wish to do is determine the strength of the electric field at a distance of 75.0 centimeters from the source. Now there are a couple of options here. I could use that same equation I've just used, plug in my value of big Q and my value of distance of 0.750 meters and solve for E, or I could recognize that this distance of 75 centimeters is three times the distance mentioned in the original problem of Part A of 25 centimeters. Now if you make the distance three times bigger, you're going to make the electric field strength go down by a factor of nine. So I can take one-ninth of the original electric field value calculated in Part A, and that should give me the electric field value here for Part C. That comes out to be 229.357 blah, blah, blah, Newtons per Coulomb, and I can simply round that to 229 Newtons per Coulomb.