Two point charges +3.0nC and −2.0nC are separated by 0.40m. Point P forms an equilateral triangle with the two charges, so P is 0.40m from each charge. Determine the magnitude and direction of the net electric field at P.
E=1.49×102N C−1, directed 19.1∘ above the horizontal toward the negative charge.
Worked solution
Each field has magnitude E=k∣q∣/r2.
E+=0.402(8.99×109)(3.0×10−9)=168.6N C−1,E−=0.402(8.99×109)(2.0×10−9)=112.4N C−1.
Because the triangle is equilateral, each field is at 60∘ to the horizontal. The positive charge field points up-right; the negative charge field points down-right.
Ex=(168.6+112.4)cos60∘=140.5N C−1,Ey=(168.6−112.4)sin60∘=48.7N C−1.E=Ex2+Ey2=148.7N C−1,θ=tan−1(48.7/140.5)=19.1∘.
Calculates field from positive charge.
[1 mark]
Calculates field from negative charge.
[1 mark]
Resolves fields into components with correct directions.
[2 marks]
Calculates resultant magnitude.
[1 mark]
Calculates and states direction.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
