Question

2. ABCD is a square of side 0.4 m. Charges of 1.6 nC, -1.6 nC and 3.2 nC are placed at the corners A, C and D respectively. Find the electric field at the point B.​

Answer 1

Answer:

AC=BD=(0.2)2+(0.2)2

=2×0.2=0.28m

∴AO=BO=CO=20.28m=0.14m

V i.e, Potential at O;Vo=4πε01[AOq1+BOq2+COq3]

V i.e, Potential at D;VD=4πε01[ADq1+BDq2+CDq3]

Work done =q[Vo