Answer:
-0.6 mJ
Explanation:
We first find the change in electric potential ΔV from
ΔV = -∫E.dr = -Ercosθ were E = magnitude of electric field = 250 N/C and r = distance of charge from the origin = √[(20 - 0)² + (50 - 0)²] = √2900 = 53.58 cm = 0.536 m. The direction is θ = tan⁻¹(50/20) = 68.2 which is the angle between E and r.
So ΔV = -Ercosθ = - 250 N/C × 0.536 m × cos68.2 = -49.77 V
The electrical potential energy ΔU = qΔV were q = care = 12 μC = 12 × 10⁻⁶ C
ΔU = qΔV = 12 × 10⁻⁶ C × -49.77 V = - 597.2 × 10⁻⁶ J = - 0.5972 mJ ≅ -0.6 mJ
