# Two similarly named point charges 10 and 40 nC are located at a distance of 30 cm from each other.

**Two similarly named point charges 10 and 40 nC are located at a distance of 30 cm from each other. Determine the point at which the field strength is zero if the system is in a vacuum.**

It is necessary to find a point on a straight line connecting two charges, where the intensities created separately by each charge are equal. Equal in magnitude and opposite in direction, they will create a net tension of zero.

Let’s write the equality:

k * q1 / x ^ 2 = k * q2 / (R-x) ^ 2,

where k is the coefficient, q1 = 40 nC, q2 = 10 nC are charges, R is the distance between charges, x is the distance from charge q1 to the desired point.

40 nC / x ^ 2 = 10 nC / (0.3 m-x) ^ 2.

After the cut:

4 / x ^ 2 = 1 / (0.3-x) ^ 2

4 * (0.3-x) ^ 2 = x ^ 2

4 (0.09-0.6x + x ^ 2) = x ^ 2

0.36-2.4x + 4x ^ 2 = x ^ 2

3x ^ 2 – 2.4x + 0.36 = 0

x ^ 2-0.8x + 0.12 = 0

First root: x = (0.8 + root (0.64-4 * 0.12)) / 2 = 0.8 + 0.4 / 2 = 0.6 m.

The distance is greater than the distance between charges. We discard the root.

Second root: x = (0.8-0.4) / 2 = 0.2 m, (does not contradict the condition).

R-x = 0.3-0.2 = 0.1 m.

Answer: The tension will be zero at a distance of 0.2 m (20 cm) from a 40 nC charge and 0.1 m (10 cm) from a 10 nC charge.