The Electric Field of a Continuous Distribution of Charge

As for the other scenarios, you'll need to use the appropriate equations for calculating the electric potential at point P.
  • #1
seto6
251
0
part 2 The Electric Field of a Continuous Distribution of Charge

Homework Statement


find the electric potential at point p
29619jm.jpg

Homework Equations



v=Kq/r...v=Er.

The Attempt at a Solution


2m3gt1d.jpg

using this pic above and dQ=(Q/L)(dX)
v=(K)(dQ)/(L-(X_i)+d) then sub dq=dX(Q/L)

we get...
(KQ/(L))(dX/(L+d-(X_i)) then integrating

V=(KQ/L)((-ln(d))+(ln(L+d)))

is this correct? if not could some one tell me where i went wrong.
thanks in advance.
 
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  • #2
That looks about right except that I would combine the logarithm arguments and write it as

[tex]V=\frac{kQ}{L}ln \left(1+\frac{L}{d}\right)[/tex]

Then you can see more clearly what happens when d is much larger than L.
 
  • #3
i see what you are doing but as d>>L it goes to zero..

but does not tell anything about what's below.

a) an infinitely long wire with total charge Q
b) an infinitely long wire with total charge Qd/L
c) a point charge of magnitude Q
d) an electric dipole with moment QL
 
  • #4
seto6 said:
i see what you are doing but as d>>L it goes to zero..

Not quite, as [itex]d\to\infty[/itex] [itex]V[/itex] will go to zero. For [itex]d\gg L[/itex] you'll want to use the Taylor series expansion of [itex]\ln(1+x)[/itex] with [itex]x=\frac{L}{d}[/itex] to get an idea of how the potential behaves.
 
  • #5


Your attempt at a solution is on the right track, but there are a few mistakes. First of all, the electric potential at point P would be the sum of the potentials due to each infinitesimal charge element, so the integral should go from 0 to L instead of L-d. Also, the equation for the electric potential due to a point charge is v=Kq/r, where r is the distance from the charge to the point at which you are measuring the potential. In this case, the distance from each infinitesimal charge element to point P would be L+d-(X_i), so that should be the denominator in your equation. Finally, when integrating, you should use the natural log function instead of the base 10 log. The correct integral would be:

V = (K/L)∫(Q/L)(dX)/(L+d-(X_i))

= (KQ/L^2)∫(dX)/(L+d-(X_i))

= (KQ/L^2)(ln(L+d-(X_i))| from 0 to L

= (KQ/L^2)(ln(L+d)-ln(d))

= (KQ/L)(ln(L+d)-ln(d))

So in summary, your attempt was mostly correct, but you made some small mistakes in the equation for the electric potential and the limits of integration. Also, remember to use natural log instead of base 10 log.
 

1. What is an electric field?

The electric field is a vector field that describes the force exerted by electric charges on other charges, either at rest or in motion.

2. How is the electric field of a continuous distribution of charge calculated?

The electric field at a point in space due to a continuous distribution of charge is calculated by summing the contributions of all the individual charges in the distribution, taking into account the distance and direction from each charge to the point.

3. Can the electric field of a continuous distribution of charge be negative?

Yes, the electric field can be negative, positive, or zero at different points in space depending on the distribution and arrangement of charges.

4. How does the electric field of a continuous distribution of charge differ from that of a point charge?

The electric field of a continuous distribution of charge is a vector field, while the electric field of a point charge is a scalar field, meaning it only has magnitude and no direction. Additionally, the electric field of a continuous distribution can vary in strength and direction at different points, while the electric field of a point charge is always spherically symmetric.

5. What are some practical applications of understanding the electric field of a continuous distribution of charge?

Understanding the electric field of a continuous distribution is crucial in many fields, such as electrical engineering, physics, and chemistry. It can be used to analyze and design circuits, understand the behavior of atoms and molecules, and study the properties of materials.

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