Electric Fields: AP Physics 2 Exam Question
Question:
Three point charges are arranged as shown below:
Q1 (+4 μC) --------------- Q2 (-2 μC)
↑
**Q3** (-3 μC)
The distance between Q1 and Q2 is 0.5 meters, and the distance between Q1 and Q3 is 0.3 meters. The electric field intensity at a point P, located on a line perpendicular to the line joining Q1 and Q2, is measured as 7.7 × 10^4 N/C.
Assume that the positive direction of the electric field intensity is defined as pointing away from positive charges and towards negative charges.
Answer:
To calculate the electric field intensity at point P due to Q1, we can use the formula for the electric field intensity:
where E is the electric field intensity, K is the electrostatic constant (9 × 10^9 Nm²/C²), Q is the charge, and r is the distance from the charge to the point where the electric field is being measured.
Plugging in the values:
To calculate the electric field intensity at point P due to Q2, we can use the same formula as above, but change the sign of the charge Q since it is negative:
To calculate the electric field intensity at point P due to Q3, we can again use the same formula:
To find the resultant electric field intensity at point P due to all three charges, we can simply add the electric field intensities:
where to 3, representing Q1, Q2, and Q3.
The magnitude and direction of the resultant electric field intensity can be found by using the Pythagorean theorem and the inverse tangent:
Note: The direction indicates the angle with respect to the positive x-axis in counterclockwise direction.
Note: Evaluating the given equations would provide the specific values based on the question details and calculations would give the final answer.