Answer:
15.38×10⁷ N/C towards the negative charge
Explanation:
From the diagram below,
E = E₁ + E₂ .................... Equation 1
Where E = Electric field at the mid point, E₁ = Electric field due to -12 µC, E₂ = Electric field due to +14.00 µC
E₁ = kq/r² .................... Equation 1
Where q = charge, r = distance, k = constant of proportionality.
Given: q = 12 µC = 12×10⁻⁶ C, k = 9×10⁹ Nm²/C² , r = 3.9 cm = 0.039 m.
Substitute into equation 1
E₁ = 9×10⁹( 12×10⁻⁶)/0.039²
E₁ = 7.1×10⁷ N/C
Similarly,
E₂ = kq'/r².................... Equation 2
Where q' = the second charge.
Given: k =9×10⁹ Nm²/C, q' = 14×10⁻⁶ C
E₂ = 9×10⁹(14×10⁻⁶ )/0.039²
E₂ = 8.28×10⁷ N/C
Therefore,
E = (7.1×10⁷)+(8.28×10⁷) = 15.38×10⁷ N/C towards the negative charge
Hence the magnitude of the electric field at the midpoint = 15.38×10⁷ N/C towards the negative charge
