Question

The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible

Answer 1

Answer:

$d_2=3.16cm$

Explanation:

So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.

In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:

$\sum{F_{x}}=0$

so:

[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]

Since $F_{13x}=F_{14x}$ we can simplify the equation as:

$-F_{12}+2F_{13x}=0$

we can now solve this for $F_{12}$ so we get:

$F_{12}=2F_{13x}$

Now we can substitute with the electrostatic force formula, so we get:

$k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta$

We can cancel $k_{e}$ and $q_{1}$

so the simplified equation is:

$\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta$

From the given diagram we know that:

$cos \theta = \frac{d_{1}}{r_{13}}$

so when solving for $r_{13}$ we get:

$r_{13}=\frac{d_{1}}{cos\theta}$

and if we square both sides of the equation, we get:

$r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}$

and we can substitute this into our equation:

$\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta$

so we can now solve this for $r_{12}$ so we get:

$r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}$

which can be rewritten as:

$r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}$

and now we can substitute values.

$r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}$

which solves to:

$r_{12}=6.16cm$

now, we must find $d_{2}$ by using the following equation:

$r_{12}=d_{1}+d_{2}$

when solving for $d_{2}$ we get:

$d_{2}=r_{12}-d_{1}$

when substituting we get:

$d_{2}=6.16cm-3cm$

so:

$d_{2}=3.16cm$