### Electric Field Due to Uniformly Charged DiskA uniformly charged disk with radius \( R = 45.0 \) cm and uniform charge density \( \sigma = 8.10 \times 10^{-3} \) C/m\(^2\) 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\[ \boxed{\hspace{2cm}} \text{ MN/C} \]#### (b) \( z = 10.0 \) cm\[ \boxed{\hspace{2cm}} \text{ MN/C} \]#### (c) \( z = 50.0 \) cm\[ \boxed{\hspace{2cm}} \text{ MN/C} \]#### (d) \( z = 200 \) cm\[ \boxed{\hspace{2cm}} \text{ MN/C} \]### Explanation:To calculate the electric field at each point, one must use principles of electrostatics, taking into account the distribution of charge over the disk and the distances specified. Specific formulas or steps should be utilized to find the electric field values. Note: The boxes are there to be filled in with the calculated values of the electric field at each specified position in the normal direction.

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A uniformly charged disk with radius R = 45.0 cm and uniform charge density o = 8.10 × 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 = 45.0 cm and uniform charge density o = 8.10 × 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

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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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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

### Electric Field Due to Uniformly Charged Disk

A uniformly charged disk with radius \( R = 45.0 \) cm and uniform charge density \( \sigma = 8.10 \times 10^{-3} \) C/m\(^2\) 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
\[ \boxed{\hspace{2cm}} \text{ MN/C} \]

#### (b) \( z = 10.0 \) cm
\[ \boxed{\hspace{2cm}} \text{ MN/C} \]

#### (c) \( z = 50.0 \) cm
\[ \boxed{\hspace{2cm}} \text{ MN/C} \]

#### (d) \( z = 200 \) cm
\[ \boxed{\hspace{2cm}} \text{ MN/C} \]

### Explanation:
To calculate the electric field at each point, one must use principles of electrostatics, taking into account the distribution of charge over the disk and the distances specified. Specific formulas or steps should be utilized to find the electric field values.

Note: The boxes are there to be filled in with the calculated values of the electric field at each specified position in the normal direction.

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