Question

Two particles are fixed to an x axis: particle 1 of charge q1​=2.07×10−8 C at x=28.0 cm and particle 2 of charge q2​=−3.61q1​ at x=62.0 cm. At what coordinate on the x axis is the electric field produced by the particles equal to zero? Number Units

Answer 1

Approximately
`(-9.78)\; {\rm cm}` (
`(-88/9) \; {\rm cm}`.)

Step-by-step explanation:

At a distance of
`r`, the electric field
`E` resulting from point charge
`q` would be
`E = k\, q / (r^(2))`, where
`k` is the Coulomb constant.

When two point charges are placed near each other, the resultant electric field would be the vector sum of the field from each charge.

Let
`x` denote the coordinate where the resultant fields is
`0`. At this position, the distance from
`q_(1)` would be
`|x - 28.0|\; {\rm cm}`, while the distance from
`q_(2)` would be
`|x - 62.0|\; {\rm cm}`.

Assume that either
`x < 28.0\; {\rm cm}` or
`x > 62.0\; {\rm cm}`, such that both
`q_(1)` and
`q_(2)` are on the same side of this position. The two electric fields would originate from the same side of this position. For the resultant electric field to be zero, set the sum of the two electric fields to be zero:

`\displaystyle (k\, q_(1))/((x - 28.0)^(2)) +(k\, q_(2))/((x - 62.0)^(2)) = 0`.

Since
`q_(2) = (-3.61)\, q_(1)`:

`\displaystyle (k\, q_(1))/((x - 28.0)^(2)) +(k\, (-3.61)\, q_(1))/((x - 62.0)^(2)) = 0`.

Simplify and solve for
`x`:

`\displaystyle (1)/((x - 28.0)^(2)) +(-3.61)/((x - 62.0)^(2)) = 0`.

The two roots are
`x = (-88/9)\; {\rm cm} \approx (-9.78)\; {\rm cm}` and
`x = (1152/29)\; {\rm cm} \approx 39.7\; {\rm cm}`. However, only
`x = (-88/9) \; {\rm cm} \approx (-9.78)\; {\rm cm}` is valid under the assumption that either
`x < 28.0\; {\rm cm}` or
`x > 62.0\; {\rm cm}`. A positive point charge at
`x \approx 39.7\; {\rm cm}` (between
`q_(1)` and
`q_(2)`) would be attracted towards
`q_(2)` while simultaneously repelled from
`q_(1)`, such that the resultant force would be non-zero.

Assume that
`28.0\; {\rm cm} < x < 62.0\; {\rm cm}`, such that this position is between
`q_(1)` and
`q_(2)`. The electric fields would originate from two different sides of this position. In the equation, the sign of one of the two fields would need to be flipped since the two fields are from two opposite directions:

`\displaystyle (k\, q_(1))/((x - 28.0)^(2)) + \left(- (k\, q_(2))/((x - 62.0)^(2))\right) = 0`.

Simplify and solve for
`x`:

`\displaystyle (1)/((x - 28.0)^(2)) +(3.61)/((x - 62.0)^(2)) = 0`.

Since both fractions are greater than zero, no real solution exists in this scenario. In other words, the electric field is non-zero when
`28.0\; {\rm cm} < x < 62.0\; {\rm cm}`.

Therefore, the only position where the electric field is zero would be
`x = (-88/9) \; {\rm cm} \approx (-9.78)\; {\rm cm}`.