Consider a point 0.5 mm above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/mV/m . What is the magnitude of the total electric field due to both charges at this location?
Question
Consider a point 0.5 mm above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/mV/m . What is the magnitude of the total electric field due to both charges at this location?
Solution
The electric field due to a single charge is given as 18 V/m. Since the electric field is a vector quantity, it has both magnitude and direction.
In this case, we have two identical positive charges. The electric field at the midpoint above the two charges due to each charge will be in the direction away from the charge (since they are positive charges).
Since the point is exactly in the middle of the two charges, the direction of the electric field due to each charge will be in opposite directions. Therefore, the total electric field at that point due to both charges will be the vector sum of the electric fields due to each charge.
However, since the electric fields due to each charge are equal in magnitude and opposite in direction, they will cancel each other out.
Therefore, the magnitude of the total electric field due to both charges at this location is 0 V/m.
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