To find the field at a point due to an arrangement of charges --- in fact, all electric fields arise due to some arrangement of charges --- we find the vector sum of the individual fields:
Enet→=∑iEi→ Net Electric Field
Electric fields are used more frequently than gravitational ones because there are two types of charge, which makes electric force and potential energy harder to keep track of than their gravitational counterparts. To apply this approach to gravitational forces --- that is, to find a net gravitational field --- one needs to repeat the steps above, with mass in place of charge (left for the reader).
Question: For the diagram above, draw (qualitatively) the electric field vectors at the points shown using the test charge method.
Answer: We will start with Test Charge 1. Test charges are always positive and have magnitude 1. Therefore we know that the test charge will want to go toward the negative charge and away from the positive charge (like charges repel and opposite charges attract). The strength of the electric field felt by the test charge is dependent on the inverse square of the distance of the charges as shown by the equation
The farther away from the source of the field, the weaker the field becomes. Therefore Test Charge 1 will experience a stronger field from the 1C charge. Because the distance from Test Charge 1 to the −1C is only slightly longer than the distance from Test Charge 1 to the 1C charge, the vectors will be similar in length. Once we have determined the relative scale of each vector, we can add them using the parallelogram method. The resultant vector is the electric field at that point.
Finding the electric field at Test Charge 2 will involve all of the same steps. First we must determine which charge Test Charge 2 is closer to. Like Test Charge 1, Test Charge 2 is closer to the 1C charge. However, Test Charge 2 is drastically closer whereas Test Charge 1 was only slightly closer. Therefore, the electric field that Test Charge 2 experiences as a result of the 1C charge will be strong, thus resulting in a longer arrow. The distance between the −1C and Test Charge 2 is large and therefore the electric field experienced by Test Charge 2 as a result of the −1C charge will be small. The resultant vectors will look something like this.