Question

At a particular moment, three small charged balls, one negative and two positive, are located as shown in the figure. Q₁ = 1 nC, Q₂ = 8 nC, and Q₃ = -3 nC.

1cm 1 cm ト..ㅓ Q3

(a1) What is the electric field at the location of Q₁, due to Q₂?

(a3) What is the net electric field at the location of Q₁, due to Q₂ and Q₃?

(b) Use the electric field you calculated above to find the force on Q₁ due to Q₂ and Q₃.

(c1) What is the electric field at location A, due to Q₁?

Answer 1

Given data Q_1 = 1nc, Q_2 = 8nc, Q_3 = -3nc a1) Electric field due to Q_2 at Q_1 is E = 1/4pielementof0 q/d^2 q = charge d = distance 1/4pielementof0 = 9 times 10^9 Nm^2/c Q_2 = 8nc d = 4cm = 4 times 10^-12 m E_Q2 = 9 times 10^9 times 8 times 10^-9/(4 times 10^-2) ^2 = 9 times 10^9 times 8 times 10^-9/16 times 10^-4 E_Q2 = 4.5 times 10^4 N/c a3) net electric field due to Q2 and Q3 at Q1 E_2 = 4.5 times 10^4 N/c E_3 = 1/ 4pielementof0 Q3/d^2 E_3 = 9 times 10^9 times 3 times 10^-9/(5 times 10^-2) ^2 = 9 times 10^9 times 3 times 10^-9/25 times 10^-4 = 27 times 10^4/25 E_3 = 10800 N/c = 1.08 times 10^4 N/c \r\n Net field due to Q2 and Q3 E_2 = 4.5 times 10^4 N/c cosTheta = adj.side/hyd E_3 = 1.08 times 10^4 N/c cosTheta = 4/5 but angle between E_2 and E_3 is 180 - Theta cosTheta = 0.16 Theta = 80.79 E_net = squareroot E_2^2 + E_3^2 + 2E_2.E_3 cos(180 - Theta) (By using parallelogram law of vector E_net = squareroot(4.5 times 10^4) ^2 + (1.08 times 10^4) ^2 + 2(4.5 times 10^4 times 1.08 times 10^4) cos(180 - 80.79) E_net = 10^4squareroot20.25 + 1.1664 + (9.72 times -0.1600) E_net = 10^4squareroot20.25 + 1.1664 - 15.12 E_net = 10^4 squareroot 2.509^2 E_net 2.509 times 10^4 N/ c