Question

Use Coulomb's law to determine the magnitude of the electric field at points A and B in the figure due to the two positive charges (Q = 6.0 µC) shown.

Magnitude at point A___N/C

Magnitude at point B___N/C

Answer 1

Coulomb's Law is F = kq1q2/r^2
Electric field is F = Eq

Thus, electric field due to a point charge is equal to E = kq/r^2

We know k (constant equal to 9x10^9 Nm^2/C^2), and we know Q (6x10^-6 C).

Thus, we only need to find the radius, solve for the electric field from each charge, and add them (as electric fields are additive).

POINT A:
first find distance (r) between left charge and A using pythag thm:
r = sqrt(10^2+5^2) = 11.18cm = 0.1118m
Note that the distance between the right charge and A is similarly r = 0.1118m.

Thus, the Electric field strength from EACH ONE of the charges is:
E(left) = (9x10^9)*(6x10^-6)/0.1118^2 = 4.32x10^6 N/C
E(right) = 4.32x10^6 N/C

HOWEVER, we must add these together to get:
E = 8.64x10^6 N/C (or 8.64 MC)


For point B:
First find the distance r(left) and r(right) using pythag:
r(left) = sqrt(5^2+5^2) = 7.07107 cm = 0.0707 m
r(right) = sqrt(15^2+5^2) = 15.81 cm = 0.1581 m

Next, we will find the individual E fields due to each charge:
E(left) = (9x10^9)*(6x10^-6)/0.0707^2 = 1.08x10^7 N/C
E(right) = (9x10^9)*(6x10^-6)/0.1581^2 = 2.16x10^6 N/C

So, adding these together gives:
E = 1.296x10^7 N/C (or 12.96 MC)