A point charge Q = -335 nC and two unknown point charges, q1 and q2, are placed

Question

A point charge Q = -335 nC and two unknown point charges, q1 and q2, are placed as shown in the figure. The electric field at the origin O, due to charges Q, q1 and q2, is equal to 434 N/C directed at an angle 39 from the negative y-axis in the fourth quadrant. The distances of charges from the origin are r = 2.12 m, r1 = 1.4 m, and r2 = 1.5 m.

On your solution paper:

1) Use E1 , E2, EQ, E--> as notations for the electric fields of charges q1 ,q2 , Q , and net field respectively.

2) draw and label the electric field vectors EQ and E--> at the origin.

3) Show the solution of the questions below neatly, explaining your work step by step. The solution starts with a physics law/equation used in terms of the symbols shown, then numerical substitutions, and final answer boxed.

4) What is the magnitude and the sign of charge q1 ? Explain the choice of the sign of the charge.

5) What is the magnitude and the sign of charge q2 ? Explain the choice of the sign of the charge.

6) What is the electric potential of the three charges at the origin?

Explanation / Answer

EQ will be in direction from origin to Q, E1 from q1 to O, and E2 from origin towards q2, E at 39 degree from negative y axis in 4th quandrant

4] Y component of field, Ey = - kQ/r^2 sin 30 degree - kq1/r1^2

-434 cos 39 degree = -9e9*-335e-9/2.12^2 *0.5 - 9e9*q1/1.4^2

q1 =  [-434 cos 39 degree +9e9*-335e-9/2.12^2 *0.5]/[-9e9/1.4^2] = 1.465*10^-7 C

= 146.5 nC

It has to be positive because if negative, it will make positive y directionof net electric field which should be not.

5] x component of field, Ex = kQ/r^2 cos 30 degree - kq2/r2^2

434 sin 39 degree = 9e9*-335e-9/2.12^2 *0.866 - 9e9*q2/1.5^2

q2 =  [434 sin 39 degree -9e9*-335e-9/2.12^2 *0.866]/[-9e9/1.5^2] = -2.135*10^-7 C

= -213.5 nC

It has to be negative because if positive, it will make x direction of net electric field negative which should not be.

6] V = summation kq/r. = 9e9*[-335e-9/2.13 + 146.5e-9/1.4 - 213.5e-9/1.5]

= -1755 V

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