Need Help understanding Part D and Part E: A point charge Q = -375 nC and two un

Question

Need Help understanding Part D and Part E:

A point charge Q = -375 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 490 N/C directed at an angle 24 ? from the negative y-axis in the fourth quadrant. The distances of charges from the origin are r =  2.46 m, r1 =  1.3 m, and r2 =  1.51 m.

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

Part A

Part complete

What is the magnitude of the electric field due to the point charge Q at the origin O?

557 NC

Correct

Part B

Part complete

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

270

Correct

Part C

Part complete

What is the magnitude and the sign of charge q1 ?

137

Answer Requested - 0 pts.

Part D

Part incomplete

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

?

Part E

Part incomplete

What is the magnitude and the sign of charge q2 ?

negative 142

Incorrect; Try Again; 5 attempts remaining

557 NC

Explanation / Answer

Part D

We know that as charge Q is negativlne charged, so it's field will will be in negative x and positive y axis( if we resolved the net electric field of Q)

And it is given that the net electric field because of all the charges is, in 4 quadrat.

That means, in positive x axis and negative y axis

So, field due to Q2 will be along the positive x axis.

That means, it will be 0° from +x axis

Part E

Net electric field's x component= field due to q2- x component of field of Q

490sin24=E-557cos30

E=681.67N/C

Now electric field is ,

E=kq/r 2

681.67=9*109*q/1.512

q=172.69 nC and it will be a negative charge.

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