Question

A positive point charge q is placed on the +y-axis at y = a, and a negative point charge –q is placed on the –y-axis at y = -a. A negative point charge –Q is located at some point on the +x-axis.

a. Draw a free-body diagram showing the forces that act on the charge –Q.
b. Find the x and y components of the net force that the two charges q and –q exert on –Q. (Your answer should involve only k, q, Q, a, and the x-coordinate of the charge -Q)
c. What is the net force on the charge –Q when it is at the origin ( x = 0 )?

Answer 1

a) See free-body diagram in attachment

b)
`F_x=0, F_y=\frac{2akqQ}{(a^2+x^2)^{(3)/(2)}}`

c)
`F=(2kqQ)/(a^2)`

Step-by-step explanation:

a)

The free-body diagram of the situation can be found in attachment.

We have:

- The charge of +q is located at y = a

- The charge of -q is located at y = -a

- The charge -Q is located at x = a

We observe that:

- The force exerted by charge +q on charge -Q is attractive, since the two charges have opposite sign, so it points to the north-west direction

- The force exerted by charge -q on charge -Q is repulsive, since the two charges have same sign, so it points to the north-east direction

So, the net force points north.

b)

We start by evaluating the situation on the x-direction first.

We observe that:

- The two charges +q and -q have same magnitude

- Also, they are located at exactly same distance from charge -Q

This means that the x-components of the force that each charge exerts on -Q are equal, but opposite in direction: therefore, they cancel each other, so the net force on the x-direction is zero:

`F_x = 0`

Instead, we observe that the y-components of the force that each charge exerts on -Q are both upward, therefore they add together.

The distance between charge +q and charge -Q is:

`r=√(a^2+x^2)`

where x is the location of the charge -Q.

The force between any of the two charges q, -q and charge -Q is given by (magnitude):

`F=(kqQ)/(r^2)=(kqQ)/(a^2+x^2)`

However, we are only interested in the y-component of the force, which is given by

`F_y = F cos \theta`

where
`\theta` is the angle between
`F_y` and
`F`. By using trigonometry,

`cos \theta=(a)/(r)=(a)/(√(a^2+x^2))`

So, substituting into the equation for
`F_y`, we find

`F_y = (kqQ)/(a^2+x^2)\cdot (a)/(√(a^2+x^2))=\frac{akqQ}{(a^2+x^2)^{(3)/(2)}}`

However, the net force is the sum of the forces due to the 2 charges, so

`F_y=\frac{2akqQ}{(a^2+x^2)^{(3)/(2)}}`

c)

Here we want to find the net force on the charge -Q when this is located at the origin, so when

x = 0

We already said that the net force on the x-direction is always zero, so

`F_x=0`

Instead the net force on the y-direction is given by

`F_y=\frac{2akqQ}{(a^2+x^2)^{(3)/(2)}}`

Therefore, by substituting x = 0, we find:

`F_y=\frac{2akqQ}{(a^2+0^2)^{(3)/(2)}}=(2akqQ)/(a^3)=(2kqQ)/(a^2)`

And the net force is in the upward direction, towards charge +q.