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Problem 20.64 An electron is at the origin, and an ion with charge +3eis at z = 17nm An electron s at the onigin. and an ion with chaige

Answer 1

The electric field of a point charge can be determined via the equation:
E = k*q/r²

Since both charges are positive, the point with zero electric field is between the two charged particles where:
k*q1/r1² = k*q2/r2²

Since the two particles are 17 nm apart:
r2 = 17 nm - r1

Thus the equation after canceling out k becomes:
q1/r1² = q2/(17 - r1)²
q1/q2 = r1² / (17 - r1)²

Plug in known values:
3e/e = r1² / (17 - r1)²
3 = r1² / (17 - r1)²
(17 - r1)² = r1² / 3

Take the square root of both sides (note the ± sign due the square root)
17 - r1 = ±r1 / 1.732
29.44 - 1.732*r1 = ±r1
29.44 = 1.732*r1 ± r1

r1 = 10.78 nm or 40.22 nm

Since r1 is within 17 nm, 10.78 nm is the answer for r1.
r2 = 17 - 10.78 =6.22 nm