Question

(vi) Determine the electric potential at the midpoint of the line joining two charges 2 × 10^5C
and -1 × 10^5 C placed in vacuum 10 cm apart

Answer 1

Answer:

Let p be the point of zero potential be p at distance r from the charge 1..

d=16 cm

Electric potential, v=q

1

/4πϵ

o

r+q

2

/4πϵ

o

(d−r)

for v=0,

r

q

1

=−

d−r

q

2

⇒

r

5×10

−8

=

0.16−r

−3×10

−8

r=10 cm

If, point p is outside,

v=q

1

/4πϵ

o

r+q

2

/4πϵ

o

(r−d)

so here, r=40 cm