Physics 12 - Quiz 2
Electric Fields and Forces
4. a) When determining the electric field at any point, we must first determine the electric field due to each point charge, and note its direction. In the diagram below E1 represents the electric field due to charge +2Q, and E2 the electric field due to charge +Q
Since we've already determined the vector direction we now calculate the magnitude of the electric field using the absolute values for the respective charges. In this case each charge is already positive.
The vectors are subtracted since they point in opposite directions.
b) The magnitude of force that each particle exerts on the other is the same however, the direction is different (opposite each other).
d) The work required from an external agent to move the +Q within 0.5 m of the +2Q charge is given by,
e) To find the velocity of the +Q when it's released and passes through the original position of 1.0 m, we need simply recognize that the change in electric potential energy will be ; a decrease in electric potential energy. A decrease in electric potential energy must mean there has been an increase in kinetic energy (i.e. conservation of energy law). Therefore since we find that
5. The solution to this problem begins with the initial recognition of the direction of each the electric fields due to each charge, as shown below. The charges have been numbered 1, 2, and 3 to reflect the corresponding electric field.
We readily see that E1 and E2 will have the same magnitude, since the charge values and radial distances to point P are the same. The magnitude of E3 will be smaller than that of E1 and E2 , since the radial distance to point P is the diagonal of the square, having a value of .
The geometry of the problem allows us see that the resultant electric field would simply be the difference between the vector sum of E1 and E2, and E3, the more formal approach of using vector components will used.
b) The force that would be exert on a -2Q charge at point P, is simply given by F = Eq
The negative sign tells us that the direction of the force is opposite that of the electric field, therefore the force acts an angle of 450 above the horizontal and to the left.
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