# The charges 90 nC and 10 nC are located at a distance of 4 cm from each other. Where should

**The charges 90 nC and 10 nC are located at a distance of 4 cm from each other. Where should the third charge be placed to keep it in balance?**

Given:

q1 = 90 nC;

q2 = 10 nC;

r = 4 cm;

Find: x.

Decision:

1) For convenience, we will translate all the variables into more convenient calculus systems. Thus, we get:

q1 = 90 nC = 9 * 10 ^ -8 C;

q2 = 10 nC = 10 ^ -8 C;

r = 4 cm = 4 ^ 10 ^ -2 m.

2) Let’s make two equations.

F13v = -F23v (in – vector); F13 = F23.

Let us denote by x the distance from the charge q1 to the charge q3. Then (r-x) is the distance from q2 to the charge q3.

3) Let’s write the formula.

F13 = k * (| q1 || q3 | / (r-x) ^ 2 and F23 = k * (| q2 || q3 | / (r-x) ^ 2

4) Let’s equate them, since F13 = F23.

k * (| q1 || q3 | / x ^ 2 = k * (| q2 || q3 | (r-x) ^ 2).

Also | q1 | / x ^ 2 = | q2 | / (r-x) ^ 2;

Let’s multiply crosswise.

| q1 | (r-x) ^ 2 = | q2 | x ^ 2;

| q1 | r ^ 2-2 | q1 | rx + | q1 | x ^ 2 = | q2 | x ^ 2;

Let’s finally open the brackets.

(| q1 | – | q2 |) x ^ 2 – 2 | q1 | rx + | q1 | r ^ 2 = 0.

From here we get an equation from which we can find x.

x = r / | q1 | – | q2 | * (| q1 | + -root of | q1q2 |).

Let’s find x1 and x2.

x1 = 4 * 10 ^ -2 / 9 * 10 ^ -8 – 10 ^ -8 * (9 * 10 ^ -8 + root of 9 * 10 ^ -8 * 10 ^ -8) = 6 cm.

x2 = 4 * 10 ^ -2 / 9 * 10 ^ -8 – 10 ^ -8 * (9 * 10 ^ -8 – the root of 9 * 10 ^ -8 * 10 ^ -8) = 3 cm.

x1 is not suitable, since in this case both forces F13 and F23 will be co-directional.

Answer: q3 should be located 3 cm from q1 and 1 cm from q2