Answer to Question #155621 in Electric Circuits for janice

Question

You have an equilateral triangle, and it happens you place three charges 28C, -52C, 60C on each vertex, the equilateral triangle is separated by the distance 44fm. What is the total net force experience by q2 due to other two charges if the charge is place at the middle vertex of the triangle?

Answer 1

In this question, there is asked to determine what total force acts on the 2nd charge, i.e., on a charge with a value of -52C.

Let's take the numbers according to the charges. For example, the 1st charge is F$_1$, and so on.

The 1st and 2nd, 2nd and 3rd charges are mutually charged signals and the force between them is directed towards each other. And, conversely, charges 1 and 3 interact in opposite directions because they have the same sign.

From this we find the forces acting on the 2nd charge:

As you can see in the picture, 2nd charge is affected by the F$_{1,2}$ and F$_{2,3}$ forces. Firstly, need to find these forces by this formula:

$F=k\times\frac{q_1\times q_2}{r^2};$

Here distance is given in the femtometer. Should covert to meter:

$1fm=1.0E-15 \ \ m =1\times10^{-15}$

$F_{1,2}=8.98\times10^9\times\frac{28\times58}{(44 \times 10^{-15})^2}=7.53\times10^{39};$

$F_{1,2}=8.98\times10^9\times\frac{58\times60}{(44 \times 10^{-15})^2}=16.14\times10^{39};$

We need to find the total of these two forces. Here we use the cosine theorem, which is used when there are two sides of a triangle and an angle between them:

$F_T=\sqrt{ F_{1,2}^2+F_{2,3}^2-2\times F_{1,2}\times F_{1,2} \times cos(a)};$

$F_{1,2}^2+F_{2,3}^2-2\times F_{1,2}\times F_{1,2} \times cos(a)$ =$(7.53\times10^{39})^2+(16.14\times10^{39})^2-$

$-2\times 7.53\times10^{39}\times16.14\times10^{39}=$ $(56.70+260.49-243.06)\times10^{78}=74.13\times10^{78};$

$F_T=\sqrt{74.13\times10^{78}}=8.61\times10^{39};$

Result: $8.61\times10^{39};$