Question

Two charges are placed on the x axis. One of the charges (q1 = +9.4 µC) is at x1 = +2.7 cm and the other (q2 = -24 µC) is at x2 = +8.9 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 = +6.2 cm. (Use the sign of your answer to indicate the direction along the x-axis.)

Answer 1

a)

Electric field
E = F/Q = kQ/r²
in Newtons/coulomb OR volts/meter
k = 1/4πε₀ = 8.99e9 Nm²/C²

field from q1 =
(spacing is 6.2–0=6.7 cm)
E1 = k(9.4*10^-6) / (0.067)² = 1.88*10⁷ V/m
direction is +x

field from q2 =
(spacing is 8.9-0=8.9 cm)
E2 = k(-24*10^-6) / (0.089)² = -2.72*10⁷ V/m
direction is –x

Net electric field = E1 +E2 =1.88*10⁷ V/m -2.72*10 V/m = - 2.53*10⁸ V/m

Direction is negative of x i.e. (- x)

b) Electric field
E = F/Q = kQ/r²
in Newtons/coulomb OR volts/meter
k = 1/4πε₀ = 8.99e9 Nm²/C²

field from q1 =
(spacing is 6.2–2.7=3.5 cm)
E1 = k(9.4*10^-6) / (.035)² = 6.9*10⁷ V/m
direction is +x

field from q2 =
(spacing is 8.9-6.2=2.7 cm)
E2 = k(-24*10^-6) / (.027)² = -2.96*10⁸ V/m
direction is –x

Net electric field = E1 +E2 =6.9*10⁷ V/m - 2.96*10⁸ V/m = - 2.27*10⁸ V/m

Direction is negative of x i.e. (- x)