What are the magnitude and direction of the electric field midway between two point charges, −12.0 µC and +14.00 µC, which are 7.800 cm apart? Give your answer using scientific notation.

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

dang that balls went far