Question

Two small beads having charges q₁ and q₂ of the same sign are fixed at the opposite ends of a horizontal insulating rod of length d. The bead with charge q₁ is at the origin. As shown in the figure below, a third small, charged bead is free to slide on the rod.

Answer 1

Force of charge 1 on bead = Force of charge 2 on bead

The formula for electric force is as follows:


F=(ke * q1*q2)/ r^2

where ke=8.988*10^9 N*m^2/C^2

and r is the distance between the two charges.

you will find r for each equation by looking at the diagram.

r will equal x for charge 1 and bead, and (d-x) for charge 2 and bead.

When we write the forces equal to each other we get: (we will call the third bead q3)

(ke * q1*q3)/ x^2 = (ke * q2*q3)/(d-x)^2

Rearrange the equation to solve for x.

upon doing so notice that ke and q3 cancel out leaving you with:

q1/x^2 = q2/(d-x)^2

continue to simplify and get x on one side:

q1/q2 = x^2/(d-x)^2

square root of (q1/q2) = x/(d-x)

square root(q1/q2)*(d-x) = x

square root(q1/q2)*(d) - square root(q1/q2)*(x) = x

square root((q1/q2)*(d) = x + square root((q1/q2)*(x)

square root((q1/q2)*(d) = x(1 + square root((q1/q2))

Hence x = square root((q1/q2)*(d) divided by (1 + square root((q1/q2))

Answer 2

Basically to solve this problem you need to first understand that in order for the bead to be in equilibrium the forces acting on both sides must be equal. So begin by writing:

Force of charge 1 on bead = Force of charge 2 on bead

The formula for electric force is as follows:


F=(ke * q1*q2)/ r^2

where ke=8.988*10^9 N*m^2/C^2

and r is the distance between the two charges.

you will find r for each equation by looking at the diagram.

r will equal x for charge 1 and bead, and (d-x) for charge 2 and bead.

When we write the forces equal to each other we get: (we will call the third bead q3)

(ke * q1*q3)/ x^2 = (ke * q2*q3)/(d-x)^2

Rearrange the equation to solve for x.

upon doing so notice that ke and q3 cancel out leaving you with:

q1/x^2 = q2/(d-x)^2

continue to simplify and get x on one side:

q1/q2 = x^2/(d-x)^2

square root of (q1/q2) = x/(d-x)

square root(q1/q2)*(d-x) = x

square root(q1/q2)*(d) - square root(q1/q2)*(x) = x

square root((q1/q2)*(d) = x + square root((q1/q2)*(x)

square root((q1/q2)*(d) = x(1 + square root((q1/q2))

Hence x = square root((q1/q2)*(d) divided by (1 + square root((q1/q2))

Answer 3

[REMOVED]

Answer 4

Two small beads having charges q1 and q2 of the same sign are fixed at the opposite ends of a horizontal insulating rod of length d. The bead with charge q1 is at the origin. A third small, charged bead is free to slide on the rod. At what position x is the third bead in equilibrium?


d/((q1/q2)^(1/2)+1)
if its wrong flip the q1 and q2 around k.