Question

The rod forms a 10/22 of circle with radius R and produces an electric field of magnitude Earc at its center of curvature P.

Answer 1

The question is about calculating the electric field at a point due to a uniformly charged arc, which is a common problem in high school level physics, specifically in the study of electrostatics.

Step-by-step explanation:

The student's question pertains to the calculation of the electric field at the center of curvature due to a uniformly charged rod that is bent into an arc of a circle. When dealing with such configurations, the electric field contribution from each differential element of charge needs to be calculated and then integrated over the entire arc. This requires knowledge of electrostatics principles such as Coulomb's Law and the concept of linear charge density. Additionally, the question involves understanding the effect a given charge distribution has on the electric field at a point of interest. This is typically tackled using techniques such as integration and considering symmetry aspects of the charge distribution.

Answer 2

`(E_q)/(E_a)=(\pi)/(2)`

Step-by-step explanation:

Assuming this problem: "Part (a) of the figure attached shows a non-conducting rod with a uniformly distributed charge +Q. The rod forms a half circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (Figure b), by what factor is the magnitude of the electric field at P multiplied?"

On this case the charge density is given by this formula:

`\lambda =(Q)/(\pi R)` assuming a half circle

We can find the force acting on the x axis with this:

`dF_x = k \int (dq)/(R^2) cos \theta = (K)/(R^2)\int 2R d \theta cos \theta`

`dF_x= - (K)/(R^2) ((Q)/(\pi R)) R \int_(-\pi/2)^(\pi/2) cos \theta d \theta`

We can cnvert the integral using the symmetrical property:

`dF_x = (KQ)/(\pi R^2) 2 \int_(0)^(\pi/2) cos \theta d \theta`

And we can find the electric field like this:

`E_(a)=(2KQ)/(\pi R^2)`

And the electric field just by the charge is given by:

`E_q = (KQ)/(R^2)`

And if we find the ratio for the two electrical fields we got:

`(E_q)/(E_a)=(\pi)/(2)`