Two point charges lie on the x axis. a charge of + 2.30 pc is at the origin, and a charge of − 4.50 pc is at x=−11.0cm. what third charge should be placed at x=+26cm so that the total electric field at x=+13.0cm is zero? express your answer to three significant figures and include appropriate units.

r₁ = distance of point A from charge q₁ = 0.13 m

r₂ = distance of point A from charge q₂ = 0.24 m

r₃ = distance of point A from charge q₃ = 0.13 m

Electric field by charge q₁ at A is given as

E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C     towards right

Electric field by charge q₂ at A is given as

E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C    towards left

Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction

Electric field at A will be zero when

E₁ = E₂ + E₃

1.225 = 0.703 + E₃

E₃ = 0.522 N/C

Electric field by charge "q₃" is given as

E₃ = k q₃ /r₃²

0.522 = (9 x 10⁹) q₃/(0.13)²

q₃ = 0.980 x 10⁻¹² C = 0.980 pC

answer: b.

explanation:

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge