AP Physics 2 Exam Question:

Two point charges, Q1 and Q2, are placed 4 meters apart. Q1 has a charge of +2 microcoulombs and Q2 has a charge of -5 microcoulombs. Calculate the electric field at a point located halfway between the two charges.

(Assume the electric constant, k, is equal to 9 x 10^9 N m^2/C^2)

Answer:

To calculate the electric field at a point located halfway between the two charges, we can use the principle of superposition and the formula for electric field due to a point charge:

Electric field magnitude, E, at a distance r from a point charge Q is given by:

E = k * |Q| / r^2

where k represents the electric constant, |Q| represents the magnitude of the charge, and r represents the distance from the charge.

In this case, we have two charges, Q1 and Q2, and we are interested in calculating the electric field at their midpoint, which is located 2 meters from each charge.

First, let's calculate the electric field due to Q1 at the midpoint:

E1 = k * |Q1| / r1^2

Plugging in the given values:

E1 = (9 x 10^9 N m^2/C^2) * (2 x 10^-6 C) / (2 m)^2

E1 = 9 x 10^9 N m^2/C^2 * 2 x 10^-6 C / 4 m^2

E1 = 4.5 x 10^3 N/C

Similarly, let's calculate the electric field due to Q2 at the midpoint:

E2 = k * |Q2| / r2^2

Plugging in the given values:

E2 = (9 x 10^9 N m^2/C^2) * (5 x 10^-6 C) / (2 m)^2

E2 = 9 x 10^9 N m^2/C^2 * 5 x 10^-6 C / 4 m^2

E2 = 11.25 x 10^3 N/C

Now, to find the net electric field, we can add the contributions of E1 and E2:

E_net = E1 + E2

E_net = 4.5 x 10^3 N/C + 11.25 x 10^3 N/C

E_net = 15.75 x 10^3 N/C

Therefore, the electric field at the point halfway between the two charges is approximately 15.75 x 10^3 N/C.

Note: The direction of the electric field is the direction of the force that a positive test charge would experience if placed at that point. However, since the charges are opposite in sign, the net force on a positive test charge at the midpoint would be zero, and hence the electric field vector points towards the negative charge.