Question

Three charged particles form a triangle: particle 1 with charge Q1 = 80.0 nC is at xy coordinates (0, 3.00 mm), particle 2 with charge Q2 is at (0, −3.00 mm), and particle 3 with charge q = 18.0 nC is at (4.00 mm, 0). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) −80.0 nC?

Answer 1

Answer:

(a)  $F_{3}=(0.829N)i^{.}$

(b) $F_{3}=-(0.621N)j^{.}$

Explanation:

For part (a)

Charge Q₁=+80×10⁻⁹C is on the y axis at y=0.003 m

And

Charge Q₂=+80×10⁻⁹C is on the y axis at y=-0.003 m

The force on particle 3 (which has charge of q=+18×10⁻⁹C) is due to vector sum of repulsive force from Q₁ and Q₂ as:

F₃₁+F₃₂=F₃

where

$|F_{31}|=k\frac{q_{3}q_{1}}{r_{31}^2} \\and\\|F_{32}|=k\frac{q_{3}q_{2}}{r_{32}^2}$

By using Pythagorean theorem r₃₁=r₃₂=0.005m

In magnitude angle notation the indicated vector addition becomes:

$F_{3}=(0.518<-37^{o} )+(0.518<37^{o})\\F_{3}=0.829<0^o$

Therefor the net force is:

$F_{3}=(0.829N)i^{.}$

For Part (b)

Switching the sign of Q₂ amount to reversing the direction of its force on q.We have:

$F_{3}=(0.518<-37^o)+(0.518<-143^o)\\F_{3}=(0.621<-190^o)$  

Therefor the net force is:

$F_{3}=-(0.621N)j^{.}$