Two charges (Q1 = +35µC and Q2 = +20µC) are fixed in place. Q1 is located at (5,

Question

Two charges (Q1 = +35µC and Q2 = +20µC) are fixed in place. Q1 is located at (5,0) cm and Q2 is located at (21,0) cm. Another charge, q = +1nC, is placed at (15,0) cm.

What is the net electric force on the third charge, q? N along the x-axis. (Note that a positive answer means along the +x axis; a negative answer means along the -x axis).

What is the net electric force on the third charge, q, if q=+2nC? N along the x-axis.

What is the net electric force on the third charge, q, if q=+4nC? N along the x-axis.

Calculate the ratio of force-to-charge for each situation above. What do you get in each case? N/C.

Use your answer above to quickly find the electric force on a charge of -60nC placed at that position. N along the x-axis.

Explanation / Answer

net electric force is F = q*E

E is the net electric field at (15,0) cm

E = E1+ E2

E1 is the field due to Q1

E2 is the field due to Q2

E1 = k*Q1/r1^2 = (9*10^9*35*10^-6)/(0.1^2) = 3.15*10^7 N/C along +X-axis

E2 = k*Q2/r2^2 = (9*10^9*20*10^-6)/(0.06^2) = 5*10^7 N/C along -X-axis

E = E1+ E2

E = (3.15-5)*10^7 = -1.85*10^7 N/C

Net force on q = +1 nC is F1 = q*E = 1*10^-9*(-1.85*10^7) = -1.85*10^-2 N/C

Net force on q = +2 nC is F2 = q*E = 2*10^-9*(-1.85*10^7) = -3.7*10^-2 N/C

Net force on q = + 4nC is F3 = q*E = 4*10^-9*(-1.85*10^7) = -0.074 N/C

Force to charge ratio is Electic field E = -1.85*10^7 N/C

in each case E is remains same E = -1.85*10^7 N/C

Force on -60 nC is F = q*E = -60*10^-9*(-1.85*10^7) = 1.11 N along +X-axis

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