Two charges, q1 and 92 are fixed on the x axis. q, has a charge of +8 µC and is located at x1= -4 cm and q2 has a charge of -10 µC and is located at x2=10 cm. a) What general area could a proton be located and experience no net force? We'll call this position xo. Choose from
Two charges, q1 and 92 are fixed on the x axis. q, has a charge of +8 µC and is located at x1= -4 cm and q2 has a charge of -10 µC and is located at x2=10 cm. a) What general area could a proton be located and experience no net force? We'll call this position xo. Choose from
College Physics
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Author:Raymond A. Serway, Chris Vuille
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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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Okay, i understand 95% of the solution to this question in regards to zero net force. On top of that, i understand most of the algebra. Where i get lost is the last part of the algebra when the professor goes from. (sqrt Iq1/q2)(x2-xo) =x1-xo to the final step before he plugs in the numbers. If you can help me understand that algebra step i would really appreciate it.
Transcribed Image Text:**Zero Net Force Solution**
Two charges, \( q_1 \) and \( q_2 \), are fixed on the x-axis. \( q_1 \) has a charge of \( +8 \, \mu C \) and is located at \( x_1 = -4 \) cm, and \( q_2 \) has a charge of \( -10 \, \mu C \) and is located at \( x_2 = 10 \) cm.
a) What general area could a proton be located and experience **no net force**? We'll call this position \( x_0 \). Choose from:
- \( x_0 < x_1 \)
- \( x_1 < x_0 < x_2 \)
- \( x_2 < x_0 \)
b) What is the value of the position \( x_0 \), where a proton would feel no net force from the charges?
![The image contains a derivation and calculation related to electric forces and equilibrium positions of charges. Here's the transcription and explanation:
---
**Derivation of Equilibrium Point**
To find the point where the forces from two charges \( q_1 \) and \( q_2 \) balance, we start with the equation:
\[ F_1 = F_2 \]
This implies:
\[ k \frac{|q_1|}{r_1^2} = k \frac{|q_2|}{r_2^2} \]
Simplifying, we have:
\[ \frac{|q_1|}{r_1^2} = \frac{|q_2|}{r_2^2} \]
Breaking it down further:
\[ \frac{|q_1|}{(x_2 - x_0)^2} = \frac{|q_2|}{(x_1 - x_0)^2} \]
Taking square roots and inverting terms:
\[ \sqrt{\frac{|q_1|/|q_2|}}(x_2 - x_0) = x_1 - x_0 \]
Solving for \( x_0 \):
\[ x_0 = \frac{x_1 - \sqrt{|q_1|/|q_2|}x_2}{1 - \sqrt{|q_1|/|q_2|}} \]
### Calculation for a Specific Example:
Using the formula derived for \( x_0 \):
\[ x_0 = \frac{-4 \, \text{cm} - \sqrt{8 \, \mu \text{C} / 10 \, \mu \text{C}}(10 \, \text{cm})}{1 - \sqrt{8 \, \mu \text{C} / 10 \, \mu \text{C}}} = -123 \, \text{cm} = -1.23 \, \text{meters} \]
### Notes:
- When taking the square root in the derivation, intermediate steps were simplified because of the squaring on both sides.
- \( x_0 \) is the equilibrium position where the forces due to charges \( q_1 \) and \( q_2 \) are equal.
- The final numerical result for this specific example places \( x_0 \) at -1.23 meters](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44c34136-e117-491b-b58f-d0d94a4d60dd%2F1f23ccc3-920f-4ca6-8a74-32bec47c7922%2Fjj2wjuo_processed.png&w=3840&q=75)
Transcribed Image Text:The image contains a derivation and calculation related to electric forces and equilibrium positions of charges. Here's the transcription and explanation:
---
**Derivation of Equilibrium Point**
To find the point where the forces from two charges \( q_1 \) and \( q_2 \) balance, we start with the equation:
\[ F_1 = F_2 \]
This implies:
\[ k \frac{|q_1|}{r_1^2} = k \frac{|q_2|}{r_2^2} \]
Simplifying, we have:
\[ \frac{|q_1|}{r_1^2} = \frac{|q_2|}{r_2^2} \]
Breaking it down further:
\[ \frac{|q_1|}{(x_2 - x_0)^2} = \frac{|q_2|}{(x_1 - x_0)^2} \]
Taking square roots and inverting terms:
\[ \sqrt{\frac{|q_1|/|q_2|}}(x_2 - x_0) = x_1 - x_0 \]
Solving for \( x_0 \):
\[ x_0 = \frac{x_1 - \sqrt{|q_1|/|q_2|}x_2}{1 - \sqrt{|q_1|/|q_2|}} \]
### Calculation for a Specific Example:
Using the formula derived for \( x_0 \):
\[ x_0 = \frac{-4 \, \text{cm} - \sqrt{8 \, \mu \text{C} / 10 \, \mu \text{C}}(10 \, \text{cm})}{1 - \sqrt{8 \, \mu \text{C} / 10 \, \mu \text{C}}} = -123 \, \text{cm} = -1.23 \, \text{meters} \]
### Notes:
- When taking the square root in the derivation, intermediate steps were simplified because of the squaring on both sides.
- \( x_0 \) is the equilibrium position where the forces due to charges \( q_1 \) and \( q_2 \) are equal.
- The final numerical result for this specific example places \( x_0 \) at -1.23 meters
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