1. A point charge q1 = -7.o µC is at the point x = - 0.400 m, y = - 0.300 m and a second point charge q2 = - 3.0 µC is at the post x = - 0.400 m, y = 0 m. Calculate the magnitude and direction of the net electric force exerted by q1 and q2 on the third charge, q3 = - 4.0 nC located at the origin.
2. In problem #1, find the total electric field at the origin.
To solve both problems, we first need to find the electric force and electric field due to each individual charge and then sum them up.
1. To find the net electric force exerted by q1 and q2 on q3, we will use Coulomb's Law, which states that the electric force between two point charges is given by:
F = (k * |q1| * |q2|) / r^2
Where F is the force, k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
Let's calculate the force exerted by q1 on q3:
Distance between q1 and q3: r1 = sqrt((-0.4 m)^2 + (-0.3 m)^2) = 0.5 m
F1 = (8.99 x 10^9 N m^2/C^2) * (7 x 10^-6 C * 4 x 10^-9 C) / (0.5 m)^2
F1 = 207.84 N repulsive force (since the charges have opposite signs)
Now, let's calculate the force exerted by q2 on q3:
Distance between q2 and q3: r2 = sqrt((-0.4 m)^2 + 0^2) = 0.4 m
F2 = (8.99 x 10^9 N m^2/C^2) * (3 x 10^-6 C * 4 x 10^-9 C) / (0.4 m)^2
F2 = 675 N attractive force (since the charges have the same sign)
To find the total force, we can use vector addition:
Fnet = F1 + F2 = 207.84 N - 675 N = -467.16 N
The magnitude of the net force is 467.16 N, and the direction is towards the negative x-axis (since both forces have components in the negative x-direction).
2. To find the total electric field at the origin, we need to find the electric field due to each charge and then sum them up.
The electric field due to a point charge is given by:
E = (k * |q|) / r^2
Where E is the electric field, k is the electrostatic constant, q is the magnitude of the charge, and r is the distance from the charge.
Let's calculate the electric field due to q1 at the origin:
Distance from q1 to the origin: r1 = sqrt((-0.4 m)^2 + (-0.3 m)^2) = 0.5 m
E1 = (8.99 x 10^9 N m^2/C^2) * (7 x 10^-6 C) / (0.5 m)^2
E1 = 251.28 N/C away from q1
Now, let's calculate the electric field due to q2 at the origin:
Distance from q2 to the origin: r2 = sqrt((-0.4 m)^2 + 0^2) = 0.4 m
E2 = (8.99 x 10^9 N m^2/C^2) * (3 x 10^-6 C) / (0.4 m)^2
E2 = 843.75 N/C towards q2
To find the total electric field, we can use vector addition:
Etotal = E1 + E2 = 251.28 N/C + 843.75 N/C = 1095.03 N/C away from q1
The magnitude of the total electric field at the origin is 1095.03 N/C, away from q1.