Electrostatic Practice Q4) Three point charges form a right-angled triangle. The

Question

Electrostatic Practice Q4) Three point charges form a right-angled triangle. Their charges are 01-4 nC, 92-6 nC and Q3-3 nC. The distance between Q1 and Q2 is 5 × 10-2 m and the distance between Q1 and Q3 is 3 × 10-2 m. What is the net electrostatic force on Q1 due to the other two charges if they are arranged as shown? (Answer: 1.48 × 10 4 N acting at an angle of 54.25° to the negative x-axis) Q5) Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased? (Answer: 0.2) Q3 =-3 nC 3 × 10-2 m Qa = +6 nC Qi = +4 nC 5×10-2 m 15

Explanation / Answer

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3.

From Coloumbs law ,force of attraction or repulsion between two charges is given by

F=KQ1Q2/r2

X-Component of net force on Q1 is

Fx=-K|Q1||Q2|/r122 =-(9*109)*(4*10-9)(6*10-9)/0.052=-8.64*10-5 N

Y--Component of net force on Q1 is

Fy=K|Q1||Q3|/r132 =(9*109)*(4*10-9)(3*10-9)/0.032=1.2*10-4 N

Magnitude

|F|=sqrt(Fx2+Fy2)=1.48*10-4 N

Direction

o=tan-1(Fy/Fx)=-54.25o

o=54.25o to the negative x-axis

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