Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by E = (1/ (x0q/( 2,,2) Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as = (1/ )(x0 2. We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain E = (x0/ 24 Evaluating the integral will lead us to Qxo E = 1 4 TTEGR? Xo (x3 + R²)/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/ E0)
Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by E = (1/ (x0q/( 2,,2) Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as = (1/ )(x0 2. We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain E = (x0/ 24 Evaluating the integral will lead us to Qxo E = 1 4 TTEGR? Xo (x3 + R²)/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/ E0)
Related questions
Question
Transcribed Image Text:Problem
Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and total charge Q) at xo distance from its center? (Consider that the
surface of the plate lies in the yz plane)
Solution
A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q.
There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring.
So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by
E = (1/
)(xOq)/(
2472)
Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as
= (1/
)(x0
24
We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain
E = (x0/
24
Evaluating the integral will lead us to
Qxo
1
1
E=
4 TE,R2 | Xo
(xỗ + R?)1/²
For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to
E = Q/(
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
