How to derive magnitude of Electric field to this?

In summary, the formula for calculating the magnitude of an electric field is E = kQ/r2, where E is the electric field strength, k is the Coulomb's constant (8.99 x 109 Nm2/C2), Q is the magnitude of the charge, and r is the distance from the charge. The direction of an electric field is determined by the direction of the force that a positive test charge would experience if placed in the field. The direction of the electric field is always away from positive charges and towards negative charges. Yes, the magnitude of an electric field can be negative, indicating that the electric field is directed towards negative charges. The magnitude of an electric field decreases as the distance from the source charge increases,
  • #1
Innuendo
26
0
[Ignore]How to derive magnitude of Electric field to this?

Homework Statement


My book states that
[tex]|E| = \frac{2kQ}{d^2} = 2\pi k\sigma [/tex]
I thought k, the coulomb constant is [tex]\frac{1}{4\pi\epsilon0}[/tex]


Homework Equations





The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
  • #2
k ≈ 8.854 187 817 x 10-12 F m-1
 
  • #3

To derive the magnitude of electric field, we can use the equation \vec{E} = \frac{\vec{F}}{q}, where \vec{F} is the electric force experienced by a test charge q in the presence of an electric field \vec{E}. In this case, we can use the Coulomb's Law, which states that the magnitude of electric force between two point charges is given by \vec{F} = \frac{kQq}{r^2}, where k is the Coulomb constant, Q is the source charge, q is the test charge, and r is the distance between the two charges.

Using this, we can derive the magnitude of electric field by substituting the expression for electric force into the first equation:

|\vec{E}| = \frac{\frac{kQq}{r^2}}{q} = \frac{kQ}{r^2}

However, it seems that in your book, they have used a different form of Coulomb's Law, where k = \frac{1}{2\pi\epsilon_0}. This is known as the Gaussian form of Coulomb's Law, and it is often used in calculations involving electric fields in continuous charge distributions, such as a charged rod or sheet.

In this case, the magnitude of electric field can be derived as:

|\vec{E}| = \frac{\frac{1}{2\pi\epsilon_0}Q}{r^2} = \frac{Q}{2\pi\epsilon_0r^2}

Alternatively, if we assume that the charge is distributed uniformly over the surface of a sphere with radius d, we can use the equation for electric field from a spherical charge distribution, which is given by |\vec{E}| = \frac{Q}{4\pi\epsilon_0d^2}. This is equivalent to the equation given in your book, |\vec{E}| = 2\pi k\sigma, since \sigma = \frac{Q}{4\pi d^2}.

In summary, the magnitude of electric field can be derived using different forms of Coulomb's Law, depending on the specific situation and distribution of charges. It is important to understand the differences between these equations and when to use each one in order to accurately calculate the electric field in a given scenario.
 

Related to How to derive magnitude of Electric field to this?

1. What is the formula for calculating the magnitude of an electric field?

The formula for calculating the magnitude of an electric field is E = kQ/r2, where E is the electric field strength, k is the Coulomb's constant (8.99 x 109 Nm2/C2), Q is the magnitude of the charge, and r is the distance from the charge.

2. How do you determine the direction of an electric field?

The direction of an electric field is determined by the direction of the force that a positive test charge would experience if placed in the field. The direction of the electric field is always away from positive charges and towards negative charges.

3. Can the magnitude of an electric field be negative?

Yes, the magnitude of an electric field can be negative. This indicates that the electric field is directed in the opposite direction of the force that a positive charge would experience. In other words, the electric field is pointing towards negative charges instead of away from them.

4. How does distance affect the magnitude of an electric field?

The magnitude of an electric field decreases as the distance from the source charge increases. This is because the electric field spreads out as it moves away from the charge, resulting in a decrease in strength.

5. What is the unit of measurement for the magnitude of an electric field?

The unit of measurement for the magnitude of an electric field is Newtons per Coulomb (N/C). This unit represents the force per unit charge that a test charge would experience in the electric field.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
882
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
735
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
807
  • Introductory Physics Homework Help
Replies
26
Views
688
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
840
Back
Top