Solving Electrical Physics Problem: Find Electric Force on Point Charge

In summary, my brother asked me this question and i completely have no idea how to do cause it is a very long time. Can anyone help, or give me an idea on how to do.
  • #1
vtran0703
2
0
My brother asked me this question and i completely have no idea how to do cause it is a very long time. Can anyone help, or give me an idea on how to do


A rectangular insulating sheet carrying a negative charge Q evenly distributed over its surface is formed into a cylinder, producing a thin walled hollow object of length L and radius R.

• We wish to find the electric force on a positive point charge q, located on the axis of the cylinder at one end.


a. Draw a sketch showing the tube, the victim charge q, and an appropriate coordinate system for describing the problem. In your sketch clearly show E at the location of q.

b. Choose and then clearly indicate the charge element dq you will use for the integration in your sketch.


c. Set up the integration. That is express F as a constant times an integral over a single variable.

Thank you
 
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  • #2
welcome to pf!

hi vtran0703! welcome to pf! :wink:

it's a fairly straightforward integration problem …

find the charge on a small area dzdθ of the cylinder, find the force at E from that charge, and then integrate over the whole cylinder

show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
It's been quite a while since I saw these problem. To be honest, I wasn't good with them back then either, but I'll try, any help would be great
 
  • #4
Remember that
[tex]\mathbf{E} = \int \frac{\lambda(\mathbf{r'}) d\mathbf{l}}{|\mathbf{r}-\mathbf{r'}|^2}\hat{(\mathbf{r}-\mathbf{r'})}[/tex]
or alternatively
[tex]\mathbf{E} = \int \frac{\lambda(\mathbf{r'}) \mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3 }d\mathbf{l}[/tex]
where r is the vector from the origin to a point in space and r' is a vector from the origin to the charge. So you make all the these infinitesimally different vectors as you sweep along the charge edge, and sum them up. A good idea for an origin is one where you can exploit symmetry in the problem, what's the most symmetric origin? What is the electric field from there?
 
Last edited:
  • #5
for your question. Solving electrical physics problems requires a systematic approach and the use of mathematical equations and principles. Here is a step-by-step guide on how to solve the problem you have described:

Step 1: Understand the problem

The first step in solving any physics problem is to clearly understand what is being asked. In this case, we need to find the electric force on a positive point charge q, located on the axis of the cylinder at one end. We are given the dimensions and charge distribution of the cylinder, and we need to use this information to calculate the electric force on the point charge.

Step 2: Draw a sketch

To better visualize the problem, it is helpful to draw a sketch. This will also help us to identify the relevant variables and choose an appropriate coordinate system for our calculations. The sketch should include the cylinder, the point charge q, and an appropriate coordinate system. We should also label the electric field (E) at the location of q.

Step 3: Choose a charge element

To calculate the electric force on the point charge, we will use the principle of superposition. This means we will break down the charge distribution of the cylinder into small charge elements and then integrate their contributions to the electric force on q. In our sketch, we should choose a charge element (dq) that represents a small portion of the cylinder's charge distribution.

Step 4: Set up the integration

Now that we have chosen a charge element, we can set up the integration to calculate the electric force on q. We know that the force between two charges is given by Coulomb's Law:

F = k * (q1 * q2) / r^2

Where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. In this case, q1 is the charge element dq and q2 is the point charge q. The distance between them can be calculated using the Pythagorean theorem.

Step 5: Solve the integral

We can now solve the integral by plugging in the appropriate values for dq, q, and r. The integral should be expressed as a constant times an integral over a single variable, which will make it easier to solve.

Step 6: Calculate the electric force

Once we have solved the integral, we can calculate the electric force on the point charge q. This will be the sum of all the contributions from the charge elements throughout the cylinder
 

Related to Solving Electrical Physics Problem: Find Electric Force on Point Charge

1. What is electric force and how is it calculated?

Electric force is a type of force that exists between two charged particles. It is calculated using Coulomb's Law, which states that the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

2. How do I find the electric force on a point charge?

To find the electric force on a point charge, you will need to know the magnitude and direction of the electric field at that point. Once you have this information, you can use the equation F = qE, where F is the electric force, q is the charge of the point charge, and E is the electric field.

3. What units are used to measure electric force?

The SI unit for electric force is Newtons (N). However, it can also be expressed in terms of Coulombs (C), meters (m), and kilograms (kg) according to the equation F = qE = (C x m)/kg.

4. Can the electric force on a point charge be negative?

Yes, the electric force on a point charge can be negative. It simply indicates that the force is acting in the opposite direction of the electric field.

5. How does distance affect the electric force on a point charge?

The electric force on a point charge is inversely proportional to the square of the distance between the two charged particles. This means that as the distance between the particles increases, the force decreases and vice versa.

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