Question

The positions of ( Q_{1} ) and ( Q_{2} ) are fixed as shown. One grid unit is 0.01 m . Angles are measured anti-clockwise from the positive ( x ) (horizontal) axis. Use ( k=9 imes 10^{9} frac{mathrm{~N} mathrm{~m}^{2}}{mathrm{C}^{2}} ) for calculations. What are the x and y components of the electric field at point A ? (a) ( E_{x}=-33 mathrm{MN} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} ) (b) ( E_{x}=-33 mathrm{MN} / mathrm{C} ), and ( E_{y}=-25.7 mathrm{MN} / mathrm{C} ) (c) ( E_{x}=33 mathrm{MN} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} ) (d) ( E_{x}=33 mathrm{MN} / mathrm{C} ), and ( E_{y}=-25.7 mathrm{MN} / mathrm{C} ) (e) ( E_{x}=0 mathrm{~N} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} )

          The positions of ( Q_{1} ) and ( Q_{2} ) are fixed as shown. One grid unit is 0.01 m .
Angles are measured anti-clockwise from the positive ( x ) (horizontal) axis.
Use ( k=9 	imes 10^{9} frac{mathrm{~N} mathrm{~m}^{2}}{mathrm{C}^{2}} ) for calculations.

What are the x and y components of the electric field at point A ?
(a) ( E_{x}=-33 mathrm{MN} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} )
(b) ( E_{x}=-33 mathrm{MN} / mathrm{C} ), and ( E_{y}=-25.7 mathrm{MN} / mathrm{C} )
(c) ( E_{x}=33 mathrm{MN} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} )
(d) ( E_{x}=33 mathrm{MN} / mathrm{C} ), and ( E_{y}=-25.7 mathrm{MN} / mathrm{C} )
(e) ( E_{x}=0 mathrm{~N} / mathrm{C} ), and ( E_{y}=25.7 mathrm{MN} / mathrm{C} )

Show more…

$The positions of ( Q1 ) and ( Q2 ) are fixed as shown. One grid unit is 0.01 m . Angles are measured anti-clockwise from the positive ( x ) (horizontal) axis. Use ( k=9 imes 10^9 fracmathrm N mathrm m^2mathrmC^2 ) for calculations. What are the x and y components of the electric field at point A ? (a) ( Ex=-33 mathrmMN / mathrmC ), and ( Ey=25.7 mathrmMN / mathrmC ) (b) ( Ex=-33 mathrmMN / mathrmC ), and ( Ey=-25.7 mathrmMN / mathrmC ) (c) ( Ex=33 mathrmMN / mathrmC ), and ( Ey=25.7 mathrmMN / mathrmC ) (d) ( Ex=33 mathrmMN / mathrmC ), and ( Ey=-25.7 mathrmMN / mathrmC ) (e) ( Ex=0 mathrm N / mathrmC ), and ( Ey=25.7 mathrmMN / mathrmC )$

Added by Joel W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 21

Instant Answer

Solved by Expert Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkouxdh4Ofnmgpwkor7Leaonfpu0Ubfpua Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkokttmmj7Lscvwvlptp4Rlhbswcdg9.Wy

11/07/2024

Step 1

03 m, 0.01 m) - \( Q_2 \) is at (0.05 m, 0.01 m) - Point A is at (0.04 m, 0.03 m) Show more…

Show all steps

lock

Thanks for your feedback!

Play audio

Feedback

Powered by NumerAI

Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkouxdh4Ofnmgpwkor7Leaonfpu0Ubfpua Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkokttmmj7Lscvwvlptp4Rlhbswcdg9.Wy and 73 other subject Physics 103 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 In this problem we are given the following charge configuration on a grid.

00:07 So let us copy the exact figure for the sake of precision, which i did right over here.

00:15 So we have these two point charges, q1 and q2.

00:21 And we have this grid where one unit is equal to let's say d, 0 .01.

00:30 Meter and we are given the coulomb constant to be equal to 9 10 to power 9 newton meter square per column squared okay we are going to assume that to the right is x and upward is y and with that we will come compute the x and y components of the electric field at the given point a.

01:11 Okay, now at this point a we have two contributions for the total field.

01:18 The contribution from the first charge and from the second charge.

01:22 So let's do them one by one.

01:26 Okay, by the way, let's remember that micro is 10 to the power minus six.

01:32 Okay now for the first field field due to this one we have this field over here e1 since q1 is positive it will be pointing away from q1 so in this direction the magnitude is k q1 over the distance square let's call this r1 from here to point a r1 squared so this is the magnitude now we also need the direction the unit direction vector so the unit direction for this vector we have the x component 3 the y component 7 and it doesn't matter if it is 3 or 3d because you are going to normalize anyways divide by the norm of this vector so we have three squared plus seven squared so this is the unit vector pointing from q1 to a okay now if you plug in the numbers we are going to obtain e1 to be equal to 2 .2051 10 to the power 7 5 .13453 10 to the power 7 in units of newton's per column.

03:27 Now let's do it for the second field or the field contribution from the second charge.

03:36 So now we see that q2 is negative so the field will point toward q2.

03:49 So this is our e2.

03:53 The magnitude is k q2 over r2 squared in absolute value.

04:01 So r2 is this distance, which is equal to.

04:10 And okay.

04:11 Let's also write down r1 because i just realized that we didn't write r1...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later