What is the direction of the electric field vector due to point charge q1 at the

Question

What is the direction of the electric field vector due to point charge q1 at the origin O?

What is the magnitude and the sign of charge q1?

What is the direction of the electric field vector due to point charge q2 at the origin O?

What is the magnitude and the sign of charge q2 ?

Constants A point charge Q = 335 nC and two unknown point charges, qi and q2 are placed as shown in the figure. The electric field at the origin O due to charges Q, 41 and 92, is equal to 788 N/C directed at an angle 35° from the negative y-axis in the fourth quadrant. The distances of charges from the origin are r- 2.06 m, T11.3 m, and 21.5 m. Use E, E, E, E as notations for the electric fields of charges 1 .92.Q, and net field respectively 1 30°

Explanation / Answer

Magnitude of charge Q at the origin will be,

E=kQ/r 2

E=9*109*335*109/2.062

E=710.482N/C

The direction of field due to q1 will be along negative y axis as then only it will result in the resulatnt in fourth quadrant.

The resultant field in the y axis is,

=788cos35

=-645.5N/C

Now the field due to Q in the y axis is,

=710.482*sin30

=355.241 N/C

Now field due to q1 should be,

-645.5=355.241+E

E=-1000.741N/C

Now field will be,

E=kq1/r 2

1000.741=9*109*q1/1.32

q1=187.92nC

Now the direction of electric field due to Q2 will be in the positive x direction.

788sin35=E-710.482cos30

E=1067.27N/C

Magitude of charge will be,

=(1067.27*1.52/9) nC

=266.8175 nC

And it will be a negative charge.

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