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