Question

A 5.4-µC point charge is located at x = 1.1 m, y = 2.9 m, and a -4.3-µC point charge is located at x = 2.3 m, y = -1.9 m.

(a) Find the magnitude and direction of the electric field at x = -2.7 m, y = 1.2 m.

magnitude	kN/C
direction	°

(b) Find the magnitude and direction of the force on a proton at x = -2.7 m, y = 1.2 m.

magnitude	N
direction	°

Answer 1

a)

q₁ = 5.4 x 10^-6 C

r₁ = 1.1 i + 2.9 j

q₂ = - 4.3 x 10^-6 C

r₂ = 2.3 i - 1.9 j

r₃ = - 2.7 i + 1.2 j

r₃₁ = r₃ - r₁ = (- 2.7 i + 1.2 j ) - (1.1 i + 2.9 j ) = - 3.8 i - 1.7 j

|r₃₁| = sqrt((- 3.8 )² + (- 1.7)²) = 4.2 m

r₂₃ = r₂ - r₃ = 2.3 i - 1.9 j - (- 2.7 i + 1.2 j ) = 5 i - 3.1 j

|r₂₃| = sqrt((5 )² + (- 3.1)²) = 5.9 m

E₁ = electric field by charge q₁ = k q₁ r₃₁ /(|r₃₁| )² = (9 x 10⁹) (5.4 x 10^-6) (- 3.8 i - 1.7 j)/(4.2)²

E₁ = 2755.1 (- 3.8 i - 1.7 j) = - (1.05 x 10⁴) i - (4.7 x 10³) j

E₂ = electric field by charge q₂ = k q₂ r₂₃ /(|r₂₃| )² = (9 x 10⁹) (4.3 x 10^-6) (5 i - 3.1 j )/(5.9)²

E₂ = (1111.75) (5 i - 3.1 j) = (5.6 x 10³) i - (3.5 x 10³) j

Net e;lectric field is given as

E = E₁ + E₂

E = - (1.05 x 10⁴) i - (4.7 x 10³) j + (5.6 x 10³) i - (3.5 x 10³) j

E = - 4900 i - 8200 j

magnitude : sqrt((- 4900)² + (- 8200)²) = 9.6 x 10³ N/C

direction : 180 + tan^-1(8200/4900) = 239.13 deg counterclockwise from +X-axis

b)

q = charge on proton = 1.6 x 10^-19 C

force on proton is given as

F = q E

F = (1.6 x 10^-19) (9.6 x 10³)

F = (15.36 x 10^-16) N

direction :239.13 deg counterclockwise from +X-axis