Question:
Two point charges, Q1 and Q2, are placed 2 meters apart in a vacuum. The magnitude of Q1 is 3.20 × 10^-6 C and the magnitude of Q2 is 4.80 × 10^-6 C.
a) What is the electric force between the two charges? b) Calculate the electric field at a point located 1 meter from charge Q1. c) Determine the electric field at a point located on the perpendicular bisector of the line joining Q1 and Q2, at a distance of 1 meter from this line.
Answer:
a) To calculate the electric force between two point charges, we can use Coulomb's Law, which states that the magnitude of the electric force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * (|Q1| * |Q2|) / r^2
Where: F is the electric force k is the electrostatic constant, approximately equal to 9.0 × 10^9 N m^2/C^2 |Q1| and |Q2| are the magnitudes of the charges r is the distance between the charges
Plugging in the values given in the question:
|Q1| = 3.20 × 10^-6 C |Q2| = 4.80 × 10^-6 C r = 2 m
F = (9.0 × 10^9 N m^2/C^2) * (3.20 × 10^-6 C * 4.80 × 10^-6 C) / (2 m)^2
F = 6.912 N
Therefore, the electric force between the two charges is 6.912 N.
b) The electric field at a point due to charge Q1 can be calculated using the formula:
E = k * |Q| / r^2
Where: E is the electric field strength k is the electrostatic constant, approximately equal to 9.0 × 10^9 N m^2/C^2 |Q| is the magnitude of the charge r is the distance from the charge
Plugging in the values given in the question:
|Q| = 3.20 × 10^-6 C (charge Q1) r = 1 m
E = (9.0 × 10^9 N m^2/C^2) * (3.20 × 10^-6 C) / (1 m)^2
E = 2.88 × 10^6 N/C
Therefore, the electric field at a point located 1 meter from charge Q1 is 2.88 × 10^6 N/C.
c) To calculate the electric field on the perpendicular bisector of the line joining Q1 and Q2, we need to consider the electric field contribution from both charges. Since the charges are of the same magnitude but opposite signs, the electric fields due to each charge will have the same magnitude but opposite directions.
Let's assume that the electric field due to Q1 is directed towards the left, and the electric field due to Q2 is directed towards the right.
By symmetry, the electric field on the perpendicular bisector will have equal magnitudes from both charges.
The electric field on the perpendicular bisector due to charge Q1 can be calculated using the formula mentioned earlier:
E1 = k * |Q1| / r^2
Where: E1 is the electric field due to charge Q1 k is the electrostatic constant, approximately equal to 9.0 × 10^9 N m^2/C^2 |Q1| is the magnitude of the charge r is the distance from the charge
Plugging in the values given in the question:
|Q1| = 3.20 × 10^-6 C (charge Q1) r = 1 m
E1 = (9.0 × 10^9 N m^2/C^2) * (3.20 × 10^-6 C) / (1 m)^2
E1 = 2.88 × 10^6 N/C
Since the electric field due to Q2 will have the same magnitude but opposite direction, the net electric field on the perpendicular bisector will be zero.
Therefore, the electric field at a point located on the perpendicular bisector of the line joining Q1 and Q2, at a distance of 1 meter from this line, is zero.