Solution
At points between the charges the individual electric fields are in same direction and do not cancel. Since charge q
2
=−4.00 q
1
located at x
2
=70 cm has a greater magnitude than
q
1
=2.1×10
−8
C located at x
1
=20 cm, a point of zero field must be closer to q
1
than to q
2
.
It must be to the left of q
1
Let x be the coordinates of P,
the point where the field vanishes.
If the field is to vanish, then
(x−x
2
)
2
∣q
2
∣
=
(x−x
1
)
2
∣q
1
∣
⇒
∣q
1
∣
∣q
2
∣
=
(x−x
1
)
2
(x−x
2
)
2
Taking the square root of both sides noting that ∣q
2
∣/∣q
1
∣=4, we obtain
x−20 cm
x−70 cm
=±20
Choosing −2.0 for consistency, the value of x found to be x=−30 cm