Question

Two charges are placed on the x axis. One of the charges (q₁ = +9.0 µC) is at x₁ = +2.8 cm and the other (q₂ = -19 µC) is at x₂ = +9.9 cm.

(a) Find the net electric field (magnitude and direction) at x = 0 cm. (Use the sign of your answer to indicate the direction along the x-axis.)

(b) Find the net electric field (magnitude and direction) at x = +5.9 cm. (Use the sign of your answer to indicate the direction along the x-axis.)

Answer 1

(a.)

Electric field is given by
E = kQ/R^2
Net electric field due to given charges will be
E = E1 - E2
We know that electric field due to positive charge is away from the positive charge and towards the negative charge.
E = kq1/r1^2 - kq2/r2^2
q1 = 9.0 $\mu$C
q2 = 19 $\mu$C
r1 = (2.8 - 0) cm = 0.028 m
r2 = (9.9 - 0) cm = 0.099 m
So,
E = 9*10^9*(19)*10^-6/0.099^2 - 9*10^9*9*10^-6/0.028^2
E = -85.86*10^6 N/C

magnitude of electric field(|E|) = 85.86*10^6 N/C
Direction is towards -ve x-axis

(b)

Electric field is given by
E = kQ/R^2
Net electric field due to given charges will be
E = E1 + E2
We know that electric field due to positive charge is away from the positive charge and towards the negative charge.
E = kq1/r1^2 + kq2/r2^2
q1 = 9.0 $\mu$C
q2 = 19 $\mu$C
r1 = (5.9 - 2.8) cm = 0.031 m
r2 = (9.9 - 5.9) cm = 0.040 m
So,
E = 9*10^9*(19)*10^-6/0.040^2 + 9*10^9*9*10^-6/0.031^2
E = +191.16*10^6 N/C
Direction is towards +ve x-axis

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