### Electric Fields due to a Charged Semicircle#### Problem StatementA uniformly charged insulating rod of length 19.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of –7.50 μC.[Diagram Description: The diagram depicts a semicircular rod with center O, indicating the uniformly charged distribution along the rod.]#### Questions##### (a) Find the magnitude of the electric field (in N/C) at O, the center of the semicircle.\[ \text{Answer: } 1.17376 \times 10^7 \text{ N/C} \]##### (b) Find the direction of the electric field at O, the center of the semicircle.- [x] to the left- [ ] to the right- [ ] upward- [ ] downward- [ ] into the page- [ ] out of the page##### (c) What if? What would be the magnitude of the electric field (in N/C) at O if the top half of the semicircle carried a total charge of –7.50 μC and the bottom half, insulated from the top half, carried a total charge of +7.50 μC?\[ \text{Answer: } 3 \times 10^7 \text{ N/C} \]#### ExplanationIn this problem, a uniformly charged rod is bent into a semicircle, which results in an electric field being generated at the center of the semicircle (point O). 1. **Magnitude of the Electric Field**: Calculations using principles of electrostatics, specifically integrating the contributions of small charge elements along the semicircle, lead to determining the electric field's magnitude.2. **Direction of the Electric Field**: Considering the symmetric charge distribution and the negativity of the charge, the overall direction of the electric field at the center O is to the left.3. **Modified Charge Distribution**: If the semicircle is divided such that the top half carries a charge of –7.50 μC and the bottom half a charge of +7.50 μC, the electric field's magnitude changes due to the opposing directions of the fields generated by these charges.#### Note:The given answers are based on calculations following the principles of Coulomb’s law and the superposition of electric fields. For a detailed step-by-step solution, one

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A uniformly charged insulating rod of length 19.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of -7.50 µC. (a) Find the magnitude of the electric field (in N/C) at o, the center of the semicircle. 1.17376E7 N/C (b) Find the direction of the electric field at 0, the center of the semicircle. O to the left O to the right O upward O downward O into the page

A uniformly charged insulating rod of length 19.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of -7.50 µC. (a) Find the magnitude of the electric field (in N/C) at o, the center of the semicircle. 1.17376E7 N/C (b) Find the direction of the electric field at 0, the center of the semicircle. O to the left O to the right O upward O downward O into the page

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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 Fields due to a Charged Semicircle

#### Problem Statement
A uniformly charged insulating rod of length 19.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of –7.50 μC.

[Diagram Description: The diagram depicts a semicircular rod with center O, indicating the uniformly charged distribution along the rod.]

#### Questions

##### (a) Find the magnitude of the electric field (in N/C) at O, the center of the semicircle.

\[ \text{Answer: } 1.17376 \times 10^7 \text{ N/C} \]

##### (b) Find the direction of the electric field at O, the center of the semicircle.

- [x] to the left
- [ ] to the right
- [ ] upward
- [ ] downward
- [ ] into the page
- [ ] out of the page

##### (c) What if? What would be the magnitude of the electric field (in N/C) at O if the top half of the semicircle carried a total charge of –7.50 μC and the bottom half, insulated from the top half, carried a total charge of +7.50 μC?

\[ \text{Answer: } 3 \times 10^7 \text{ N/C} \]

#### Explanation

In this problem, a uniformly charged rod is bent into a semicircle, which results in an electric field being generated at the center of the semicircle (point O).

1. **Magnitude of the Electric Field**: Calculations using principles of electrostatics, specifically integrating the contributions of small charge elements along the semicircle, lead to determining the electric field's magnitude.
2. **Direction of the Electric Field**: Considering the symmetric charge distribution and the negativity of the charge, the overall direction of the electric field at the center O is to the left.

3. **Modified Charge Distribution**: If the semicircle is divided such that the top half carries a charge of –7.50 μC and the bottom half a charge of +7.50 μC, the electric field's magnitude changes due to the opposing directions of the fields generated by these charges.

#### Note:

The given answers are based on calculations following the principles of Coulomb’s law and the superposition of electric fields. For a detailed step-by-step solution, one

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