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Anonymous asked in Science & MathematicsPhysics · 1 month ago

What is the magnitude of the electric field created by qB at D?

Two point charges qA=-6nC & qB are placed in vacuum. The charge qA is at a distance 12cm from point D. charge qB is at distance 6cm from D. The resultant electric field vector created by the charges at D is along xaxis and makes 30 degrees with electric field A and 60 degrees with electric field B. 

1 Answer

  • 1 month ago

    Line qA-D is length d=12cm and makes angle 30º with x-axis.

    Note, qA is negative (reverses field direction) so the angle might be 180º + 30º = 210º.  The wording is not clear so I’m using 30º.

    Line qB-D is length 6cm makes angle 60º with x-axis.

    EA is the field from D to qA.

    EB is the field from qB at D.

    Resultant field lies along x-axis so the y-components of EA and EB cancel:

    EAsin(30º) + EBsin(60º) = 0

    (k.qA/d²)sin(30º) +  EBsin(60º) = 0

    Put the values in and solve for EB

    (This doesn’t use the ‘6cm’ distance given in the question.  So maybe I’ve misunderstood or maybe this distance is needed in a later part of the question.)

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