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.
- Steve4PhysicsLv 71 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.)