Field and force with three charges At a particular moment,...

Field and force with three charges At a particular moment, three small charged balls, one negative and two positive, are located as shown in the figure. Q1 = 1 nC, Q2 = 9 nC, and Q3 = -5 nC. Remember that you must first convert all quantities to S.I. units. 1 nC = 1 nanocoulomb = 1e-9 C. (a1) What is the electric field at the location of Q1, due to Q2? E2 = (0, 5.06e4, 0) N/C (a2) What is the electric field at the location of Q1, due to Q3? E3 = (0) N/C (a3) What is the net electric field at the location of Q1, due to Q2 and Q3? Enet = N/C (b) Use the electric field you calculated above to find the force on Q1 due to Q2 and Q3. F = N

Transcript

00:02 Hello students, in this question there are three charges q1, q2, q3 in a xy plane.

00:09 So q2 is at the origin and q1 is on the y -axis having coordinates 0 ,4 and q3 is on the x -axis having coordinates 3 ,0.

00:22 So here charge q1 is given as 1 nc, q2 is 9 nc and q3 is minus 5 nc.

00:31 And r12 is the distance between charge q1 and q2 that is 0 .04 meter, r23 is the distance between charge q2 and q3 that is 0 .03 meter and r13 is the distance between charge q1 and q3.

00:46 So if we take this as a right angle triangle according to pythagoras theorem, the hypotenuse will be root of base square plus altitude square that will be root of r12 square plus r23 square so that is 0 .05 meter.

01:02 So in the a part, in the a1 we have to find out the electric field experienced by charge q1 due to q2 so that can be written as e2.

01:14 So e2 will be equal to q2k divided by r12 into j cap.

01:24 So j cap is the direction.

01:26 So here e2, so e2 electric field e2 represents the electric field experienced by the charge q1 due to q2 and j cap is the direction of electric field and here k is a constant called as 1 by 4 pi epsilon 0 and that is equal to 9 into 10 raised to 9.

01:48 So we can substitute here so this will be equal to 9 into 10 raised to 9 that is charge q2 10 raised to minus 9 into k that is 9 into 10 raised to 9 divided by r12 so that is 0 .04.

02:07 So here r there is r12 square it is a square of the distance so we have to square it into j cap, j cap is the direction of electric field e2.

02:18 So if we calculate we will get the electric field e2 will be equal to 5 .06 j cap newton per meter.

02:31 So in next part we have to find out the electric field experienced by charge q1 due to q3.

02:37 So this is the direction of electric field e3.

02:42 So in next part a2 part we have to find out e3, e3 vector that is the electric field experienced by charge q1 due to q3...

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