Answer:
(a) (i) Potential,
V = VA + VB
=
2.4 × 105V.
(ii)
Electric field, E = EB ? EA
towards
1.5 charge.
(b)
(i) In this case distance AQ becomes i.e.
0.18 m.
Now
Potential, V = VA + VB
= 2
× 105 V.
(ii) Electric
Field,
Here
Where
and
=
4.167 × 105 NC?1
=
6.944 × 105 NC?1
Using
(i), (ii), and (iv),
For
direction,
we
get tan
Resultant
electric field is inclined at an angle with
AB.
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