three charges, +2.5×10^-6, -4.8×10^-6, and -6.3×10^-6, are located at (-20m,0.15m), (0.50m,-0.35m), and (-0.42m,-0.32m) respectively. What is the electric field at the origin?
To find the electric field at the origin due to the given charges, we need to calculate the contribution from each charge using Coulomb's law and then sum them up vectorially.
The electric field (E) at a point due to a point charge (q) is given by the equation:
E = k * (q / r^2)
where k is the Coulomb's constant (8.99 × 10^9 N m^2/C^2), q is the charge, and r is the distance between the point charge and the point where we want to calculate the electric field.
Now let's calculate the contribution from each charge to the electric field at the origin:
1. For the first charge (+2.5×10^-6 C) located at (-20m, 0.15m):
- Calculate the distance between the charge and the origin (r1):
r1 = sqrt((-20m)^2 + (0.15m)^2)
- Calculate the electric field contribution (E1):
E1 = k * (2.5×10^-6 C) / (r1^2)
2. For the second charge (-4.8×10^-6 C) located at (0.50m, -0.35m):
- Calculate the distance between the charge and the origin (r2):
r2 = sqrt((0.50m)^2 + (-0.35m)^2)
- Calculate the electric field contribution (E2):
E2 = k * (-4.8×10^-6 C) / (r2^2)
3. For the third charge (-6.3×10^-6 C) located at (-0.42m, -0.32m):
- Calculate the distance between the charge and the origin (r3):
r3 = sqrt((-0.42m)^2 + (-0.32m)^2)
- Calculate the electric field contribution (E3):
E3 = k * (-6.3×10^-6 C) / (r3^2)
Finally, to find the electric field at the origin, sum up the contributions from each charge vectorially:
E_Total = E1 + E2 + E3
This will give you the magnitude and direction of the electric field at the origin.