Question

A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 6000 N/C and is directed in the positive x direction. A point charge

-6.3

Answer 1

Answer 2

Okay - so the field is pointing in the + x direction which means the low potential is to the left. Now the negative charge at x = 0, y = 0, has a filed that points toward it and the charge's field has a magnitude:

Ec = kq/r^2 where k = 9x10^9 Nt-m^2/Coul^2 and q = -6.3x10^-9 coul, r = distance from the charge.

So now at x = -0.18 m, r^2 = x^2 = (-0.18 m)^2 =0.0324 m^2 and the field from the charge points toward the charge, which is in the same direction of the uniform field. So the magnitude is:

Etotal = Euniform + Ec = 6000 N/C + kq/(0.0324) = 4500 N/C +1750 N/c = 6250 N/C


Now on the y - axis r = 0.18 m r^2 = 0.0324 m^2 but now Ec points in the -y direction so the magnitude to Etotal is:

Etotal = sqrt(Euniform^2 + Ec^2) = 4828.3 N/C