What Are the Limits for Calculating the Radial Electric Field Potential?

AI Thread Summary
The discussion focuses on calculating the electric potential from the radial electric field near a point charge 'q'. The formula for the electric field is given as E = (1 / 4∏ε) * (q/r^2), and the potential difference is derived by integrating this expression. The correct limits for the integration are confirmed to be from infinity to a point r2, with V(∞) set to zero. The user initially had the limits reversed but corrected them after clarification. Understanding these limits is crucial for accurately calculating the potential.
chris_avfc
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Homework Statement



Radial component of an electric field near charge 'q'.

E = (1 / 4∏ε) * (q/r^2)

Need to find the potential of this which is done through integrating this and taking the negative answer, i.e.

ΔV = -∫E dr

This is to be along a radial path towards the charge, starting from infinity to r2.


The Attempt at a Solution



So I have

ΔV = - (q/4∏ε) ∫1/r^2 dr

As the charge is constant.
Just unsure on the limits, I'm guessing its ∞ and r2, would that be right?
 
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\Delta V=V(r)-V(\infty)=-\frac{q}{4\pi \epsilon _0}\int_{\infty}^r{\frac{1}{r^2}dr},

with V(∞)=0


ehild
 
ehild said:
\Delta V=V(r)-V(\infty)=-\frac{q}{4\pi \epsilon _0}\int_{\infty}^r{\frac{1}{r^2}dr},

with V(∞)=0


ehild

Awesome, cheers mate. I had my limits the wrong way around.
 
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