Electric Field Video Lessons

Concept

# Problem: A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00 nC is at the point x = 0.600 meters, y = 0.A) Calculate the magnitude E of the net electric field at the origin due to these two point charges.B) What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two point charges.

###### FREE Expert Solution

The electric field is expressed as:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{k}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$, where k is the Coulomb constant, r is the distance of the charge, and E is the electric field.

A)

From the given information, the distance of charge q1 from the origin is given by:

r1 = sqrt(0.6002 + 0.8002) = 1.0 m

The distance of q2 from the origin:

r2 = sqrt(0.6002 + 02) = 0.600 m

E1 = kq/r12 = (9.0 × 109)(4.00 × 10-9)/1.0= 36.0 N/C

E2 = (9.0 × 109)(6.00 × 10-9)/0.600= 150 N/C

Calculate the x and y components of the electric field at the origin due to the two charges.

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###### Problem Details

A point charge q1 = -4.00 nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00 nC is at the point x = 0.600 meters, y = 0.

A) Calculate the magnitude E of the net electric field at the origin due to these two point charges.

B) What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two point charges.