Question

Question 4 Two particles are fixed to an x axis: particle 1 of charge -3.15 x 10 C is at the origin and particle 2 of charge +3.15 x 10 C is at x2 18.5 cm. Midway between the particles, what is the magnitude of the net electric field? Number the tolerance is +/-596 Click if you would like to Show WorN/C or V or V/mtion: N/s V/m s N/C m Open Show Work

Answer 1

Q_1 = -3.15 times 10^-7 C Q_2 = + 3.15 times 10^-7 C x = x_2/2 x_2 = 18.5 times 10^-2 m E_1^bar & E_2^bar are same direction, E_xt = E_1 + E_2 E_xt = 2 k Q/x^2 = 2 times 9 times 1069 times 3.15 times 10^-7/(9.25 times 10^-2)^2 E_xt = 6.63 times 10^5 N/C Q_1/Q_2 = 4 x_5 = 31.2 cm from sr - ph E_xt = 0 at 2 x_s/3 = 20.8 cm = x_0 (at P) Therefore E_1 = E_2 = > k Q_1/(L + x_0)^2 = k Q_2/x_0^2 = > Q_1/Q_2 = (L + x_0/x_0)^2 Therefore squareroot 1/4 = L + x_0/x_0 = > 1/2 = L + x_0/x_0 = > L = x_0/2 = 10.4 cm from E_xt^bar (x) is positive maximum at 31.2 cm E_xt (x) is -negative maximum at 15.6 cm = x_1 E_xt = k [Q_1/(L + x_1)^2 - Q_2/x_1^2] = 9 times 10^9 [12 e/(26 times 10^-2)^2 - 3e/(15.6 times 10^-2)^2]