c) A positive charge q initially at rest is placed at point P. The charge is released and is pushed by the E field towards infinity. Show that the work done by the electric field is W = 2kQq √d²+x² cord
c) A positive charge q initially at rest is placed at point P. The charge is released and is pushed by the E field towards infinity. Show that the work done by the electric field is W = 2kQq √d²+x² cord
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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Use the given information to solve part C
![### Electric Field Analysis of Two Point Charges
**Problem Setup:**
Two equal positive point charges \( +Q \) are placed on the y-axis, each at distance \( d \) above and below the x-axis. We need to analyze the electric field at a point \( P \).
#### a) Electric Field at Point P
**Diagram Explanation:**
The diagram shows two charges \( +Q \) positioned symmetrically about the x-axis, with point \( P \) on the x-axis at distance \( x \) from the origin. The electric field vectors due to each charge at point \( P \) are represented, with the components of these vectors in the x and y directions.
#### b) Expression for the Electric Field at Point P
We are tasked to prove that the electric field \( \vec{E} \) at point \( P \) is given as:
\[
\vec{E} = \frac{2kQx}{(x^2 + d^2)^{3/2}} \hat{i}
\]
**Derivation:**
- **Electric Field Vector \( \vec{E} \):**
\[
\vec{E} = \vec{E}_{1} + \vec{E}_{2}
\]
where \( \vec{E}_{1} \) and \( \vec{E}_{2} \) are the electric field contributions from each charge.
- **Components of \( \vec{E} \):**
\[
\vec{E}_{x} = E_{x1} + E_{x2}
\]
\[
E_{x} = \frac{2kQ \cdot x}{(x^2 + d^2)^{3/2}}
\]
- **Using cosine relation:**
\[
\cos \theta = \frac{x}{\sqrt{x^2 + d^2}}
\]
Substituting into the component equation, we find:
\[
E_{x} = \frac{2kQx}{(x^2 + d^2)^{3/2}} \hat{i}
\]
Thus, the electric field at point \( P \) is shown to be aligned in the x-direction with magnitude \( \frac{2kQx}{(x^2 + d^2)^{3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F418fd126-7f9b-4db4-a312-2e5f96462821%2Fc0e3e287-97ce-4487-af94-4bd7f956c9d1%2Fsgqk65a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Electric Field Analysis of Two Point Charges
**Problem Setup:**
Two equal positive point charges \( +Q \) are placed on the y-axis, each at distance \( d \) above and below the x-axis. We need to analyze the electric field at a point \( P \).
#### a) Electric Field at Point P
**Diagram Explanation:**
The diagram shows two charges \( +Q \) positioned symmetrically about the x-axis, with point \( P \) on the x-axis at distance \( x \) from the origin. The electric field vectors due to each charge at point \( P \) are represented, with the components of these vectors in the x and y directions.
#### b) Expression for the Electric Field at Point P
We are tasked to prove that the electric field \( \vec{E} \) at point \( P \) is given as:
\[
\vec{E} = \frac{2kQx}{(x^2 + d^2)^{3/2}} \hat{i}
\]
**Derivation:**
- **Electric Field Vector \( \vec{E} \):**
\[
\vec{E} = \vec{E}_{1} + \vec{E}_{2}
\]
where \( \vec{E}_{1} \) and \( \vec{E}_{2} \) are the electric field contributions from each charge.
- **Components of \( \vec{E} \):**
\[
\vec{E}_{x} = E_{x1} + E_{x2}
\]
\[
E_{x} = \frac{2kQ \cdot x}{(x^2 + d^2)^{3/2}}
\]
- **Using cosine relation:**
\[
\cos \theta = \frac{x}{\sqrt{x^2 + d^2}}
\]
Substituting into the component equation, we find:
\[
E_{x} = \frac{2kQx}{(x^2 + d^2)^{3/2}} \hat{i}
\]
Thus, the electric field at point \( P \) is shown to be aligned in the x-direction with magnitude \( \frac{2kQx}{(x^2 + d^2)^{3
Expert Solution
Step 1: Draw the electric field at P
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