Question

Electric field intensity at point 'B' due to a
point charge 'Q' kept at point 'A' is 24 NC-1
and
the electric potential at point 'B' due to same
charge is 12 JC-1
. Calculate the distance AB and
also the magnitude of charge Q.

Answer 1

Answer:

0.5 m, 0.67nC

Explanation:

since E=V/d

d=V/E

=12/24

=0.5 m (=AB)

also V=kQ/d

Q=Vd/k

=12×0.5/9×10⁹

=0.67 × 10–⁹ C =0.67nC