Two point charges, -20μC and +40μC , are separated by a distance of 17.0cm .
What is the magnitude of the electric field at a point halfway between the two charges?
Express your answer to two significant figures and include the appropriate units
To find the magnitude of the electric field at a point halfway between two charges, we can use the formula for the electric field due to a point charge:
E = k * |q| / r^2
where E is the electric field, k is Coulomb's constant (8.99 × 10^9 N m^2/C^2), |q| is the magnitude of the charge, and r is the distance from the charge.
In this case, we have two charges:
- Charge 1: -20μC (negative charge)
- Charge 2: +40μC (positive charge)
The distance between them is 17.0 cm. Since we want to find the electric field at the point halfway between the charges, the distance from each charge to that point is half of the total distance, which is 17.0 cm / 2 = 8.5 cm = 0.085 m.
Now let's calculate the electric fields due to each charge separately and then add them up.
Electric field due to Charge 1:
E1 = k * |q1| / r1^2
= (8.99 × 10^9 N m^2/C^2) * (20 × 10^-6 C) / (0.085 m)^2
Electric field due to Charge 2:
E2 = k * |q2| / r2^2
= (8.99 × 10^9 N m^2/C^2) * (40 × 10^-6 C) / (0.085 m)^2
To find the total electric field at the halfway point, we add the magnitudes of the electric fields due to each charge:
E_total = |E1| + |E2|
Now we can calculate the magnitude of the electric field at the halfway point by substituting the values into the equations:
E_total = (8.99 × 10^9 N m^2/C^2) * (20 × 10^-6 C) / (0.085 m)^2 + (8.99 × 10^9 N m^2/C^2) * (40 × 10^-6 C) / (0.085 m)^2
Calculating this expression will give you the magnitude of the electric field at the halfway point between the two charges. Round the final answer to two significant figures and include the appropriate units.