Electric Field at a Specific Point due to Two Charges

Click For Summary
SUMMARY

The discussion focuses on calculating the magnitude of charge Q2 required to achieve a zero electric field at a point 1.75 meters from charge Q1, which is 17 C and located at the origin. The charges are 2.5 meters apart on the x-axis. The correct approach involves recognizing that the electric field is a vector quantity, necessitating consideration of direction. The final calculation reveals that Q2 must be -8.673 C to nullify the electric field produced by Q1 at the specified point.

PREREQUISITES
  • Understanding of Coulomb's Law
  • Familiarity with electric field concepts
  • Knowledge of vector addition in physics
  • Ability to perform algebraic manipulations
NEXT STEPS
  • Study vector addition in electric fields
  • Learn about electric field calculations involving multiple charges
  • Explore the implications of charge polarity on electric field direction
  • Review problem-solving techniques for electrostatics in Physics 2
USEFUL FOR

Students in college-level Physics courses, particularly those studying electrostatics and electric fields, as well as educators seeking to clarify concepts related to charge interactions.

Jgs19
Messages
1
Reaction score
0
Hey I'm a new time member but I hope to use this forum for my college Physics 2 class, anyways:

1. Question: Two charges are D = 2.5 m apart and lie on the x-axis. What must be the magnitude of the charge on the right one, Q2, (in C) so that the electric field is zero at x = 1.75 meters. Let Q1 be at the origin and have a charge 17 C.

2. Equations I'm using.
E(total)=E1+E2...
E=k*q/r^2

3. My attempt:
E1=(8.99E9)(17C) / (1.75^2) = 4.990E10
E2= k*Q / (1.75^2)
E(total) = E1 + E2
0 = E1+E2
-E1=E2
Q= (-4.990E10)(1.75^2) / (8.99E9)
Q= -16.9987

I've just learned electrical fields / coulombs law, for a problem like this I just need some explanation on what I'm doing wrong. I thing what the biggest problem is and confusing this is it initially is 2.5m apart but they want you to find it at 1.75m, I'm going to keep working but I feel like the 2.5 is relevant and I'm not using it. Thanks in advance.
 
Physics news on Phys.org
Jgs19 said:
Hey I'm a new time member but I hope to use this forum for my college Physics 2 class, anyways:

1. Question: Two charges are D = 2.5 m apart and lie on the x-axis. What must be the magnitude of the charge on the right one, Q2, (in C) so that the electric field is zero at x = 1.75 meters. Let Q1 be at the origin and have a charge 17 C.

2. Equations I'm using.
E(total)=E1+E2...
E=k*q/r^2

3. My attempt:
E1=(8.99E9)(17C) / (1.75^2) = 4.990E10
E2= k*Q / (1.75^2)
E(total) = E1 + E2
0 = E1+E2
-E1=E2
Q= (-4.990E10)(1.75^2) / (8.99E9)
Q= -16.9987

I've just learned electrical fields / coulombs law, for a problem like this I just need some explanation on what I'm doing wrong. I thing what the biggest problem is and confusing this is it initially is 2.5m apart but they want you to find it at 1.75m, I'm going to keep working but I feel like the 2.5 is relevant and I'm not using it. Thanks in advance.
it's quite simple if you understand the concepts...see electric field is a vector, so you should take the direction under consideration before setting the equation,According to data,Q1=+17C, so the field due to it on point(1.75,0) is towards +ve x-axis.So to nulify its effect and make field zero at (1.75,0),Q2 would be -ve and direction of field due to it would be towards -ve x-axis.Now if we only talk about the magnitude,then,...
E1=E2
so, Q1/r1^2=Q2/r2^2
so, 17/(1.75)^2=Q2/(1.25)^2
so, Q2=8.673C.
This was the magnitude that we found.Now, as we discussed earlier, Q2 has to be -ve.So,
Q2=-8.673C.
Hope this was helpful...
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
2
Views
3K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K