AP Physics 2 Exam Question: Electric Charge and Field
A point charge with a charge of +6.0 μC is positioned at the origin in a xy-plane. A second point charge with a charge of -3.0 μC is positioned at coordinates (4.0 m, 0) in the same plane. Calculate the net electric field at the point (2.0 m, 3.0 m) due to these two charges.
Answer with Step-by-Step Detailed Explanation:
To find the net electric field at the point (2.0 m, 3.0 m) due to the two charges, we need to calculate the electric field produced by each charge individually and then add them together vectorially.
The electric field, E, at any given point in space due to a point charge can be calculated using the formula:
E = k * (Q / r^2) * u
Where:
First, let's calculate the electric field produced by the charge of +6.0 μC at point (2.0 m, 3.0 m):
r1 = sqrt((2.0 m - 0)^2 + (3.0 m - 0)^2) = sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13) m
u1 = [(2.0 m - 0)/sqrt(13) m]î + [(3.0 m - 0)/sqrt(13) m]ĵ
E1 = k * (6.0 μC) / (sqrt(13) m)^2 * u1
Next, let's calculate the electric field produced by the charge of -3.0 μC at point (2.0 m, 3.0 m):
r2 = sqrt((2.0 m - 4.0 m)^2 + (3.0 m - 0)^2) = sqrt((-2)^2 + 3^2) = sqrt(4 + 9) = sqrt(13) m
u2 = [(2.0 m - 4.0 m)/sqrt(13) m]î + [(3.0 m - 0)/sqrt(13) m]ĵ
E2 = k * (-3.0 μC) / (sqrt(13) m)^2 * u2
Finally, let's calculate the net electric field vector by adding E1 and E2:
E_net = E1 + E2
Make sure to calculate the respective components of E1 and E2 using vectors î and ĵ:
E_net = (E1_x + E2_x)î + (E1_y + E2_y)ĵ
Simplify the expressions and solve for E_net to obtain the net electric field vector at point (2.0 m, 3.0 m).
Remember to provide the answer with the correct units and round to an appropriate number of significant figures.