Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
For this question, can you explain how we get the integral from dFz to Fz. Can you also explain how im supposed to know which proportion to use? In some similar question we use proportions such as dQ/dl and in other question we use dz/d theta. How do i know which proportion to use for the question?

For this question, can you explain how we get the integral from dFz to Fz. Can you also explain how im supposed to know which proportion to use? In some similar question we use proportions such as dQ/dl and in other question we use dz/d theta. How do i know which proportion to use for the question?

icon
Related questions
Question

For this question, can you explain how we get the integral from dFz to Fz. Can you also explain how im supposed to know which proportion to use? In some similar question we use proportions such as dQ/dl and in other question we use dz/d theta. How do i know which proportion to use for the question?

A thin uniformly charged ring (charge Q, radius R) is in the xy plane, as shown below. Find the force acting on charge q located at point P that is a distance z from the center of the ring. dl dF cos 0 dQ q 0 da 8 dF q dQ r² q dQ cose r² Perpendicular components of dF cancel due to the ring symmetry. r² = R² + z² and Using the proportion: dF₂ = k dQ dl F₂ = k dFz dF = k 2πR = k Z q dQ R² + z² √√R² + z² dFz ) dQ= Or use proportion = = = k cose = where dl - Rda Qdl 2πR dQ Q da 2πT 오 za 2π (R² 4 245 3/2 S = 2π z q k QRda 2πR Z da = r Q da 2π (R² + z²)3/2 = z q dQ (R²+z²)³/2 dQ 1 Απερ Z R² + z² Qda 2π = Q da 2πT zqQ 3 (R² + z²)² At the center of the ring (z = 0), the force F = 0. For z >> R, R in the denominator can be ignored and the electric field simplifies to: 1 qQ F₂ 4περ z2 (the same result as for the charge Q at the center of the ring).
Transcribed Image Text:A thin uniformly charged ring (charge Q, radius R) is in the xy plane, as shown below. Find the force acting on charge q located at point P that is a distance z from the center of the ring. dl dF cos 0 dQ q 0 da 8 dF q dQ r² q dQ cose r² Perpendicular components of dF cancel due to the ring symmetry. r² = R² + z² and Using the proportion: dF₂ = k dQ dl F₂ = k dFz dF = k 2πR = k Z q dQ R² + z² √√R² + z² dFz ) dQ= Or use proportion = = = k cose = where dl - Rda Qdl 2πR dQ Q da 2πT 오 za 2π (R² 4 245 3/2 S = 2π z q k QRda 2πR Z da = r Q da 2π (R² + z²)3/2 = z q dQ (R²+z²)³/2 dQ 1 Απερ Z R² + z² Qda 2π = Q da 2πT zqQ 3 (R² + z²)² At the center of the ring (z = 0), the force F = 0. For z >> R, R in the denominator can be ignored and the electric field simplifies to: 1 qQ F₂ 4περ z2 (the same result as for the charge Q at the center of the ring).
Expert Solution
steps

Step by step

Solved in 3 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS