Electric Field at a Specific Point due to Two Charges

In summary, the conversation discusses calculating the magnitude of a charge at a specific point in an electric field using Coulomb's law. The equations and attempts to solve the problem are also mentioned, along with a solution provided by another member. The concept of electric field direction is emphasized in finding the correct magnitude of the charge.
  • #1
Jgs19
1
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.
 
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  • #2
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...
 

1. What is an electric field?

An electric field is a region in space where electrically charged particles, such as electrons and protons, experience a force. This force can either attract or repel the charged particles, depending on their charges.

2. How is an electric field created?

An electric field is created by any object that has a net electric charge. This charge can be positive or negative, and it can be generated by the movement of electrons or protons. The strength of the electric field is determined by the magnitude of the charge and the distance from the charged object.

3. What is the formula for calculating the strength of an electric field?

The formula for calculating the strength of an electric field is E = F/Q, where E is the electric field strength, F is the force acting on a charged particle, and Q is the charge of the particle. This formula is also known as Coulomb's law.

4. How does the direction of an electric field relate to the direction of the force on a charged particle?

The direction of the electric field is defined as the direction in which a positive test charge would move if placed in the field. The direction of the force on a charged particle in an electric field is the same as the direction of the electric field at that point. This means that a positive charge will experience a force in the same direction as the electric field, while a negative charge will experience a force in the opposite direction.

5. What are some real-life applications of electric fields?

Electric fields have many practical applications in our daily lives. They are used in electronics, such as in computers and smartphones, to control the movement of electrons. They are also used in medical imaging techniques like X-rays and MRI to create images of the body. Electric fields are also used in particle accelerators, which are essential in studying subatomic particles and nuclear reactions.

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