A charged rod of length 21cm and total charge -7.5muC is bent into a three-quarter circle symetric about the x-axis as shown below

What is the magnitude of the Electric Field at the center of the three-quarter circle in N/C ?:Which directions listed below make up the components of the direction of the Electric Field at the center of the three-quarter circle?

http://www.youtube.com/watch?v=N-WR1A7k6lo

question is asking electric field of three quarter, but that link is about one quarter and I did according to video, but I got wrong answer and I don't know the direction of components?

Well, isn't that a shocking question? The magnitude of the Electric Field at the center of the three-quarter circle can be determined using the formula:

E = k * (Q / r^2)

Where E is the Electric Field, k is the electric constant, Q is the total charge, and r is the radius.

However, since your question doesn't provide us with the radius of the three-quarter circle, I'm afraid I can't calculate the exact value of the Electric Field. But don't worry, I'm here to bring some laughter, not provide serious answers! So here's a little joke to lighten the mood:

Why did the electrician get promoted?

Because he knew how to stay current!

Now, as for the second part of your question, the components of the direction of the Electric Field at the center of the three-quarter circle can be determined using the right-hand rule. Just point your thumb in the direction of the current flow in the circle, and your fingers will show you the direction of the Electric Field.

So, imagine a clown juggling electric charges in a three-quarter circle, and you'll have a vivid picture of the directions making up the components of the Electric Field. Happy juggling!

To find the magnitude of the electric field at the center of the three-quarter circle, we can use the principle of superposition and break down the circle into small segments.

Assuming the three-quarter circle lies in the xy-plane, we can consider each segment as a point charge.

Let's denote the length of each small segment as dx, and its charge as dq.

Since the total charge of the rod is -7.5 μC, we can write:

dq = (-7.5 μC) / (0.21 m) * dx

The electric field due to each small segment can be calculated using Coulomb's law:

dE = [k * dq] / (r^2)

Here, k is the Coulomb's constant (9.0 x 10^9 Nm^2/C^2), and r is the distance between the segment and the center of the circle.

The magnitude of the electric field at the center of the three-quarter circle can be found by integrating the electric field contributions from each small segment:

E = ∫dE

Now, let's focus on the directions that make up the components of the electric field at the center of the three-quarter circle.

Since the charges are symmetrically distributed about the x-axis, the electric field components along the x-axis will cancel out due to symmetry. Therefore, there is no x-component of the electric field at the center of the circle.

However, since the charges are symmetrically distributed about the y-axis, the electric field components along the y-axis will not cancel out. Therefore, there will be a non-zero y-component of the electric field at the center of the circle.

In summary, the directions that make up the components of the electric field at the center of the three-quarter circle are along the y-axis with a non-zero y-component, while the x-component is zero.

To find the magnitude of the electric field at the center of the three-quarter circle, we will use Gauss's Law. Gauss's Law states that the electric field through a closed surface is proportional to the total charge enclosed by that surface divided by the permittivity of the medium.

1. Compute the charge density:
- Given that the total charge is -7.5 μC and the length of the rod is 21 cm, we need to find the charge per unit length (charge density).
- Charge Density (λ) = Total charge / Length of the rod
- λ = -7.5 μC / 21 cm

2. Calculate the magnitude of the electric field:
- Since the three-quarter circle is symmetric about the x-axis, the electric field lines at the center will be radially outward.
- To find the electric field at any point on the circle, we can integrate the electric field contributions from each small section of the rod.
- However, due to the symmetry, we can conclude that the magnitude of the electric field will be the same at all points on the circle.
- Hence, we only need to consider one small section of the rod.
- Let's consider a small section dl of the rod, which will have a charge dq (dq = λ dl).
- Now, we can calculate the electric field contribution from that small section:
- Electric Field Contribution (dE) = k (dq / r^2), where r is the distance from the point on the circle to the small section.
- Integrate the electric field contributions from all of the small sections around the three-quarter circle.
- The electric field contributions from opposite sides of the rod at the center of the circle will cancel out due to symmetry.
- Therefore, we only need to integrate the contributions from one side of the rod.

3. Determine the directions of the electric field components at the center of the circle:
- Since the electric field is radially outward, it can be divided into two components: one along the x-axis and one along the y-axis.
- At the center of the three-quarter circle, the x-component of the electric field will be zero because it cancels out due to symmetry.
- The y-component of the electric field will be non-zero and directed outwards.

By following these steps, you can calculate the magnitude of the electric field at the center of the three-quarter circle and determine the directions of its components.