How Do You Calculate Electric Potential and Field for a Finite Line Charge?

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To calculate the electric potential for a finite line charge with density λ extending from -a to +a, the potential V at a point r on the x-axis (where r > a) is derived using the equation V = (1/4πε₀)∫(dq/r). The differential charge dQ is expressed as λdx, leading to the integration of Qdx/2a over the limits of ±a, resulting in V = Q/(4πε₀r). There is uncertainty regarding the inclusion of λ in the final expression for V, and further clarification is sought on the integration process. Additionally, the electric field can be determined using Gauss's law and the relationship E = -dV/dr. The discussion highlights the need for accurate limits and integration techniques in solving for electric potential and field.
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Homework Statement


A line charge has density λ and extends along the x-axis from -a to +a. Find the electric potential at a point r on the x-axis (r>a). Use your result to find the E-field at r.

Homework Equations


V = \frac{1}{4\pi \epsilon _0}\int\frac{dq}{r}

The Attempt at a Solution


I've said so far:
dQ = λdx = Qdx/2a

Then I made the substitution into the above equation, and integrated wrt x with limits ±a, leaving me with \frac{Q}{4\pi \epsilon _0 r}.

However I'm not sure I made the correct integration here. I also believe that the answer for V should have λ in it somewhere. Have I gone wrong somewhere?

I haven't actually gotten round to looking at the E-field yet - I'll get there once this is completed!

Any pointers would be appreciated.

Thanks
 
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Is this correct? I've spent my morning researching this, and I can't seem to find anything to compare it to - I'm a little confused as to what to do when the point is somewhere on the same line. Are my limits correct?
 
You could find the Electric field using gauss's law and then use E=-dV/dr to solve for V.
 
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