AP Physics 2 Exam Question:
A point charge of +4 μC is placed at the origin in an electric field. Another point charge of -3 μC is placed at a distance of 2 meters along the positive x-axis. Determine the electric field strength at a point located at 3 meters along the positive y-axis from the origin.
Answer:
Given data:
To find the electric field strength at the point located at 3 meters along the positive y-axis, we need to calculate the electric field due to both charges at that point and then determine the net electric field.
The electric field due to a point charge at a distance r is given by the formula:
E = k * (q / r^2)
Where:
Let's calculate the electric field due to charge q1 at the given point:
E1 = k * (q1 / r1^2)
Since q1 = +4 μC and r1 = √((0)^2 + (3)^2) = 3 m:
E1 = (9 * 10^9 N m^2/C^2) * (4 * 10^-6 C / (3 m)^2) ≈ 1.333 * 10^5 N/C
Next, let's calculate the electric field due to charge q2 at the given point:
E2 = k * (q2 / r2^2)
Since q2 = -3 μC and r2 = √((2)^2 + (3)^2) ≈ 3.6056 m:
E2 = (9 * 10^9 N m^2/C^2) * (-3 * 10^-6 C / (3.6056 m)^2) ≈ -1.862 * 10^4 N/C
Since the electric field is a vector quantity, we need to consider the direction of the electric field due to each charge separately. The electric field due to q1 points vertically upward along the positive y-axis, and the electric field due to q2 points towards the left along the negative x-axis.
Therefore, the net electric field at the given point is the vector sum of the electric fields due to q1 and q2:
E_net = E1 + E2
E_net = (1.333 * 10^5 N/C) + (-1.862 * 10^4 N/C) ≈ 1.147 * 10^5 N/C
The net electric field at the point located 3 meters along the positive y-axis from the origin is approximately 1.147 * 10^5 N/C, pointing vertically upward along the positive y-axis.