Radial Electric Field

In summary, the conversation discusses the calculation of work done by an electric field on a charge moved from point A to B, using the equation W=q∫E⋅dr. The correct value for q is determined, and the integration is performed using the boundaries of A=10 and B=4. The resulting work is found to be negative, which is expected since the field does negative work and the potential energy increases.
  • #1
Winzer
598
0

Homework Statement


Carefully study the following plot of electric field E in N/C versus distance r in m and answer the questions that follow. The electric field is directed radially outward, and the variation of E with r is independent of direction.

Refering to the figure, determine the work done by the electric field on a 7.75 C charge moved from A to B.



Homework Equations


[tex] W=q \int\vec{E}\dot\vec{dr}[/tex]

The Attempt at a Solution


First I found the charge caused by the electric field( I used E=25 and r=2 since they intersect:
[tex] q=\frac{E r^2}{\epsilon_{o}}\longrightarrow q= q=\frac{25* 2^2}{\epsilon_{o}}= 1.11E^-8
C[/tex]
-----
I am finding Work done from a to b:
[tex] W= q\int\vec{E}\dot\vec{dr}[/tex] I am pretty sure I am integrating with respect to r.
[tex] W= q\int\frac{kq_{1}}{r^2}dr[/tex] a=10 b=4
So what do you think?
 

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  • #2
anyone?
 
  • #3
Winzer said:
anyone?

Yes, looks right to me. Although you wrote [tex] q=\frac{E r^2}{\epsilon_{o}}[/tex] you meant [tex] q=\frac{E r^2}{k}[/tex] right?

Also here:
[tex] W= q\int\vec{E}\dot\vec{dr}[/tex]

[tex] W= q\int\frac{kq_{1}}{r^2}dr[/tex]

q is the 7.75C and q1 is the 1.11E-8C (doesn't matter mathematically... but for your substitution to make sense E = kq1/r^2... hence q1 is the charge creating the field)...

just making sure because you used q for 1.11E-8 before...

Everything looks good.
 
  • #4
learningphysics said:
Yes, looks right to me. Although you wrote [tex] q=\frac{E r^2}{\epsilon_{o}}[/tex] you meant [tex] q=\frac{E r^2}{k}[/tex] right?

Also here:
[tex] W= q\int\vec{E}\dot\vec{dr}[/tex]

[tex] W= q\int\frac{kq_{1}}{r^2}dr[/tex]

q is the 7.75C and q1 is the 1.11E-8C (doesn't matter mathematically... but for your substitution to make sense E = kq1/r^2... hence q1 is the charge creating the field)...

just making sure because you used q for 1.11E-8 before...

Everything looks good.
That is Correct i meant K.
So: [tex] (9.0e^9* 1.11E^-8* 7.75E^-6 * \int \frac{1}{r^2} dr[/tex] boudries of a=10 to b=4. = -1.161E-4 J, shouldn't work be postive, we are going against the e-field
Oh and by the way the charge is suppose to be 7.75 uC.
 
  • #5
Winzer said:
That is Correct i meant K.
So: [tex] (9.0e^9* 1.11E^-8* 7.75E^-6 * \int \frac{1}{r^2} dr[/tex] boudries of a=10 to b=4. = -1.161E-4 J, shouldn't work be postive, we are going against the e-field
Oh and by the way the charge is suppose to be 7.75 uC.

It should be negative since we're calculating the work done by the field, not the change in electric potential energy.

The change in electric potential energy is the negative of the work done by the field... the field does negative work... potential energy increases.
 
  • #6
Ok i get it,
Thank you
 

1. What is a radial electric field?

A radial electric field is a type of electric field that extends outwards from a central point in all directions. It is often represented by lines of force that radiate outwards from a positive charge or towards a negative charge.

2. How is a radial electric field created?

A radial electric field is created by the presence of electric charges. Positive charges will create an outward radial electric field, while negative charges will create an inward radial electric field.

3. What is the significance of a radial electric field?

A radial electric field plays a crucial role in many electrical phenomena, such as the movement of charged particles, the behavior of electric fields in capacitors, and the operation of certain electronic devices.

4. How is a radial electric field measured?

A radial electric field can be measured using an electric field meter or by calculating the electric field strength at a given point using the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the charge, and r is the distance from the charge.

5. Can a radial electric field be manipulated?

Yes, a radial electric field can be manipulated by changing the magnitude or location of the electric charges that create it. This is the basis of many electronic devices, such as capacitors and transistors, which use controlled changes in radial electric fields to control the flow of electricity.

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