What is the correct electric field vector at point P due to source charge q=-25C? Use k=1x10¹0Nm²/C². 9-25 C 4.0m O(-10¹0 10¹03) N/C O(-10¹0 10¹03) N/C 0 1 x 10¹0 N/C O(x10¹0+ × 10¹0) N/C X P 3.0m 4
What is the correct electric field vector at point P due to source charge q=-25C? Use k=1x10¹0Nm²/C². 9-25 C 4.0m O(-10¹0 10¹03) N/C O(-10¹0 10¹03) N/C 0 1 x 10¹0 N/C O(x10¹0+ × 10¹0) N/C X P 3.0m 4
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![### Understanding the Electric Field Vector Problem
#### Problem Statement:
What is the correct electric field vector at point P due to source charge q = -25C? Use k = 1x10^10 Nm²/C².
#### Diagram Explanation:
The diagram is a right-angled triangle that provides the positions of the source charge and the point where the electric field is to be determined. Here are the details of the diagram:
- The source charge \( q = -25C \) is located at one corner of the triangle (left).
- Point P is the point where the electric field is being calculated and is located at the other corner of the right triangle.
- The distance between the source charge and the base of the triangle to point P is shown as 4.0 meters horizontally (to the right) and 3.0 meters vertically (upward).
The positive direction is defined with respect to the conventional coordinate system: right as the positive x-axis, and up as the positive y-axis. The electric field produced by a negative charge points toward the charge.
#### Answer Choices:
1.
\[
\left( -\frac{3}{5} \times 10^{12} i - \frac{4}{5} \times 10^{12} j \right) \, \text{N/C}
\]
2.
\[
\left( \frac{4}{5} \times 10^{12} i - \frac{3}{5} \times 10^{12} j \right) \, \text{N/C}
\]
3.
\[
1 \times 10^{12} \, \text{N/C}
\]
4.
\[
\left( \frac{4}{5} \times 10^{12} i + \frac{3}{5} \times 10^{12} j \right) \, \text{N/C}
\]
5.
\[
\left( \frac{3}{5} \times 10^{12} i + \frac{4}{5} \times 10^{12} j \right) \, \text{N/C}
\]
To determine the correct electric field vector at point P due to source charge \( q = -25C \), the following formula for the electric field due to a point charge is applied:
\[
\vec{E} = \frac{k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a5ac3c6-e4ab-423a-91e1-ff33bdae0d07%2F26a0e267-649f-42f8-a585-2c177126d24a%2Ffxhdlr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Electric Field Vector Problem
#### Problem Statement:
What is the correct electric field vector at point P due to source charge q = -25C? Use k = 1x10^10 Nm²/C².
#### Diagram Explanation:
The diagram is a right-angled triangle that provides the positions of the source charge and the point where the electric field is to be determined. Here are the details of the diagram:
- The source charge \( q = -25C \) is located at one corner of the triangle (left).
- Point P is the point where the electric field is being calculated and is located at the other corner of the right triangle.
- The distance between the source charge and the base of the triangle to point P is shown as 4.0 meters horizontally (to the right) and 3.0 meters vertically (upward).
The positive direction is defined with respect to the conventional coordinate system: right as the positive x-axis, and up as the positive y-axis. The electric field produced by a negative charge points toward the charge.
#### Answer Choices:
1.
\[
\left( -\frac{3}{5} \times 10^{12} i - \frac{4}{5} \times 10^{12} j \right) \, \text{N/C}
\]
2.
\[
\left( \frac{4}{5} \times 10^{12} i - \frac{3}{5} \times 10^{12} j \right) \, \text{N/C}
\]
3.
\[
1 \times 10^{12} \, \text{N/C}
\]
4.
\[
\left( \frac{4}{5} \times 10^{12} i + \frac{3}{5} \times 10^{12} j \right) \, \text{N/C}
\]
5.
\[
\left( \frac{3}{5} \times 10^{12} i + \frac{4}{5} \times 10^{12} j \right) \, \text{N/C}
\]
To determine the correct electric field vector at point P due to source charge \( q = -25C \), the following formula for the electric field due to a point charge is applied:
\[
\vec{E} = \frac{k
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