Two point charges are on the x-axis: q1 = 8.00 nC at x = +16.0 m and q2 = 6.00 nC at x = -9.00 m. What is the x-component of the electric field at the origin?

• A. 0.947 N/C
• B. -0.947 N/C
• C. 0.385 N/C
• D. -0.385 N/C

Answer: (C)

The electric field from a point charge points away from the charge if the charge is positive, and its magnitude is k q / r^2 with r being the distance to the point of interest. Here all fields lie along the x-axis, so we just assign signs based on directions toward or away from each charge.

For the charge at +16.0 m (positive q), the field at the origin points toward negative x, with magnitude E1 = k q1 / r1^2 = (8.99×10^9)(8.00×10^-9)/(16.0^2) ≈ 0.281 N/C, so E1_x ≈ -0.281 N/C.

For the charge at -9.00 m (positive q), the field at the origin points toward positive x, with magnitude E2 = k q2 / r2^2 = (8.99×10^9)(6.00×10^-9)/(9.0^2) ≈ 0.666 N/C, so E2_x ≈ +0.666 N/C.

Add the x-components: E_x = E1_x + E2_x ≈ -0.281 + 0.666 ≈ +0.385 N/C.

So the x-component of the electric field at the origin is about 0.385 N/C in the +x direction.