Electric Field of a Uniformly Charged Ring

[ILoveCollege]

Homework Statement

A uniformly charged ring of radius 8.1 cm has a total charge of 118 micro Coulombs. The value of the Coulomb constant is 8.98755e9 N M^2/C^2. Find the magnitude of the electric field on the axis of the ring at 1.15 cm from the center of the ring. Answer in units of N/C.

Homework Equations

F= k Qq/ r^2
E= kq/r^2

The Attempt at a Solution

I tried subtracting 1.15 cm from 8.1 cm for "r" and plugged that "r" value in the F equation but that answer is wrong. By axis , do they mean horizontally (as in along the diameter) or vertically?

 

[ILoveCollege]

do i just do the pyth. theorem to solve for "r" ( square root of 8.1^2 + 1.15^2)?

 

[kerimek]

I would first clarify with the person who provided this problem what they mean by "axis." This term can be interpreted in different ways, and it is important to have a clear understanding in order to solve the problem correctly.

Assuming that "axis" refers to the horizontal axis (along the diameter) of the ring, I would approach the problem by first calculating the linear charge density of the ring, which is the total charge divided by the circumference of the ring. This would give us the charge per unit length.

Next, I would use the formula for the electric field due to a charged ring, which is E = kλx / (x^2 + R^2)^3/2, where λ is the linear charge density, x is the distance from the center of the ring, and R is the radius of the ring.

Plugging in the given values, we can calculate the electric field at 1.15 cm from the center of the ring. Be sure to convert all units to SI units before plugging in to the formula.

If "axis" refers to the vertical axis, the approach would be similar, but using the formula for the electric field due to a charged disk instead. This formula is E = kσz / (z^2 + R^2)^3/2, where σ is the surface charge density, z is the distance from the center of the disk, and R is the radius of the disk.

Again, I would clarify with the person who provided the problem to ensure that the correct formula is being used.