Static Electricity Legacy Problem #26 Guided Solution

An object which is charged alters the electrical properties of the space which surrounds it in such a manner that any other object placed in that space experiences the influence of that charge. We say that a charged object creates an electric field. We can calculate the strength of an electric field of a charged object if we multiply the quantity of charge the object has by the value of Coulomb's law constant, 8.99 times 10 to the 9th, Newton's meter squared per Coulomb squared, and if we divide by the distance of separation, or the distance of separation from the object that creates the electric field. Here in this question we're given that there are two charges, Q1 and Q2, both creating an electric field. And what we wish to do is to calculate the strength of the net electric field at a location exactly midway between them. That's a distance of .123 meters from either one of the charges. So this problem involves calculating two electric field vectors, one of them being the electric field vector created by charge 1, the other the electric field vector created by charge 2. I can do that in a stepwise fashion. I'm going to call the electric field created by charge 1, E1, and I'm going to calculate its magnitude by going K times this Q1 value, divided by .123 squared, and what I do, I end up getting this value for E1, 4943.935 Newton's per Coulomb. Now the direction of this electric field is the direction that a positive test charge would be pushed by Q1, or pulled by Q1, when placed at that location. Since Q1 is positive, it would push a positive test charge to the right. So E1 is directed rightward. Now when it comes to calculating electric field by charge created by charge 2, I have to use the same equation, only I'm going to substitute in for Q, the value of Q2. So I go K times the Q2 value times the distance from Q2, which is .123 meters, and I square that distance and I end up getting this number, 3589.1070 Newton's per Coulomb. Now the direction of E2, or the electric field created by charge 2, is to the left, in the direction that a positive test charge would be pushed by Q2. And since Q2 is positive, a positive test charge would be pushed away from it. So now I have to calculate the net electric field from these two magnitudes and directions, E1 and E2. Now since E1 is the bigger of the two vectors, it's going to be partially cancelled, but not fully cancelled, by the E2. So the net electric field is going to be directed to the right, in the direction of the bigger E1 value. In order to find the magnitude of this net electric field, I simply subtract the smaller from the larger. I do this whenever two vectors are going in opposite directions, and I have to add them. When I'm done, I get this numerical value, 1354.8285 Newton's per Coulomb to the right, and I can round that to two significant digits, such that the answer is 1350 Newton's per Coulomb.