two points of 2 C and a 5 C are 10 meters apart. A third point charge of 6 C is placed between them. at what distance from 2 C charge must the 6 C charge is placed, in such way that stay in equillibrium?

To find the distance from the 2 C charge where the 6 C charge must be placed for equilibrium, we can use the principle of electrostatic equilibrium. In this case, since the charges are all along a straight line, we can assume that the equilibrium position is also on the same line.

The principle of equilibrium states that the net force on a charge at equilibrium must be zero. Mathematically, this can be expressed as:

F_1 + F_2 = 0,

where F_1 is the force exerted on the 6 C charge by the 2 C charge and F_2 is the force exerted on the 6 C charge by the 5 C charge.

The electrostatic force (F) between two charges (q_1 and q_2) separated by a distance (r) is given by Coulomb's law:

F = k * (q_1 * q_2) / r^2,

where k is the electrostatic constant.

Let's calculate the forces and set them equal to each other to find the equilibrium distance:

F_1 = k * (2 C) * (6 C) / (r_1)^2,
F_2 = k * (5 C) * (6 C) / (r_2)^2.

Since the distance between the charges is given as 10 meters, we have:
r_1 + r_2 = 10 m.

Setting F_1 equal to F_2, we can solve for the unknown distance r_1:

k * (2 C) * (6 C) / (r_1)^2 = k * (5 C) * (6 C) / ((10 m - r_1)^2).

Cancelling out the electrostatic constant and the (6 C) term:

2 / (r_1)^2 = 5 / ((10 m - r_1)^2).

Now we can solve this equation to find the value of r_1 that satisfies equilibrium.