Algebra of two equal positive charges

[Lisa...]

I need some help in finding a way to solve the following problems.:

- Two equal positive charges q are on the y axis, one at y= +a and the other at y= -a.

a) Show that the electric field on the x-axis is along the x-axis with
E_x= 2kqx(x² + a²)^-3/2.

b) Show that near the origin, when the x is much smaller than a, E_x is approximately 2kqx/a³.

c) Show that for values of x much larger than a, E_x is approximately 2kq/x². Explain why you would expect this result even before calculating it.

Thanks for your efforts!

 

[inha]

You need to show your efforts before you get help. So, where do you think you should start?

 

[kyle8921]

a) To find the electric field at a point on the x-axis, we will use Coulomb's Law which states that the electric field at a point due to a point charge is given by E = kq/r^2, where k is the Coulomb's constant, q is the charge and r is the distance from the charge to the point.

In this case, we have two charges of equal magnitude q, one at y=+a and the other at y=-a. The distance from each of these charges to a point on the x-axis, say x, will be (x^2+a^2)^1/2.

Therefore, the electric field at this point due to the charge at y=+a will be E1 = kq/(x^2+a^2)^1/2 and the electric field at this point due to the charge at y=-a will be E2 = kq/(x^2+a^2)^1/2.

Since the electric fields are in the same direction, we can add them together to get the total electric field at this point: E = E1 + E2 = 2kqx(x^2+a^2)^-3/2.

b) When x is much smaller than a, we can approximate (x^2+a^2)^-3/2 to be approximately equal to a^-3, since x^2 << a^2. Therefore, the electric field at this point will be approximately E = 2kqx/a^3.

c) Similarly, when x is much larger than a, we can approximate (x^2+a^2)^-3/2 to be approximately equal to x^-2, since x^2 >> a^2. Therefore, the electric field at this point will be approximately E = 2kq/x^2.

We would expect this result because as we move further away from the charges, the effect of each charge on the electric field at a point decreases. This is because the distance between the charges and the point increases, making the value of (x^2+a^2)^1/2 or (x^2+a^2)^-3/2 smaller. As a result, the electric field at the point will decrease, and when x is much larger than a, the electric field will be dominated by the charge that is closer to the point, resulting in an electric