### Electric Field Due to a Uniformly Charged DiskA uniformly charged disk with radius \( R = 25.0 \) cm and uniform charge density \( \sigma = 7.40 \times 10^{-3} \) C/m\(^2\) lies in the \( xy \)-plane, with its center at the origin. This problem explores finding the electric field (in MN/C) due to the charged disk at various distances \( z \) along the \( z \)-axis.#### Given:- **Radius of the disk:** \( R = 25.0 \) cm- **Charge density:** \( \sigma = 7.40 \times 10^{-3} \) C/m\(^2\)#### Problem:What is the electric field (in MN/C) due to the charged disk at the following locations?**(a) \( z = 5.00 \) cm**\[ \boxed{} \text{ MN/C} \]**(b) \( z = 10.0 \) cm**\[ \boxed{} \text{ MN/C} \]**(c) \( z = 50.0 \) cm**\[ \boxed{} \text{ MN/C} \]**(d) \( z = 200 \) cm**\[ \boxed{} \text{ MN/C} \]### Explanation- **Radius \( R \) and charge density \( \sigma \)** are constants for the disk.- The electric field \( E \) at a distance \( z \) from the center of the disk, along the perpendicular axis, is what needs to be calculated.- The mentioned locations \( z \) are measured along the \( z \)-axis from the plane of the disk.### Steps to Calculate the Electric Field1. **Find the exact expression for the electric field** due to a uniformly charged disk at a point on the axis perpendicular to the disk through its center. The general formula for the electric field at a distance \( z \) from the center of a uniformly charged disk is given by: \[ E(z) = \frac{\sigma}{2 \epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right) \] Where: - \( \epsilon_0 \) is the permittivity of free space (\(

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A uniformly charged disk with radius R = 25.0 cm and uniform charge density o = 7.40 x 10-3 C/m2 lies in the xy-plane, with its center at the origin. What is the electric field (in MN/C) due to the charged disk at the following locations? (a) z = 5.00 cm MN/C (b) z = 10.0 cm MN/C (c) z = 50.0 cm MN/C (d) z = 200 cm MN/C

A uniformly charged disk with radius R = 25.0 cm and uniform charge density o = 7.40 x 10-3 C/m2 lies in the xy-plane, with its center at the origin. What is the electric field (in MN/C) due to the charged disk at the following locations? (a) z = 5.00 cm MN/C (b) z = 10.0 cm MN/C (c) z = 50.0 cm MN/C (d) z = 200 cm MN/C

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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