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Derivation of electric potential due to point charge 
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#1
Sep911, 09:57 AM

P: 62

Electric potential is the work done in moving a unit charge from infinity to a point in an electric field.
Electric potential due to point charge: [itex]V=\int \vec{E}\cdot d\vec{s}[/itex] [itex]V=\int E\cdot ds cos \vartheta[/itex] if the stationary charge is positive and if the test charge is is moved from infinity to point P then [itex]V=\int E\cdot ds cos 180[/itex] [itex]V=KQ \int \frac {1}{r^2} ds cos 180[/itex] now how to solve further as stationary charge is positive the electric field is outward i.e. from p to infinity and movement of charge is from infinity to P specially the signs and the direction of the field and the direction of ds and definite integration from P to infinity or from infinity to P? please give some imaginary picture or idea of how the test charge moves I am confused with it please help. 


#2
Sep911, 10:03 AM

P: 3,014

If the angle between E and ds is 180^{o}, then, where in what direction is ds directed?



#3
Sep911, 10:23 AM

P: 62

from infinity towards Q
now i have attached an image in the original post 


#4
Sep911, 10:24 AM

P: 3,014

Derivation of electric potential due to point charge
Yes, but how do we call this direction? It has a "special relation" to one of the variables in your equations.



#5
Sep911, 10:35 AM

P: 62

it should have relation with
[itex] \frac {1}{r^2} [/itex] 


#6
Sep911, 10:36 AM

P: 3,014

Yes. How do we call a straight line starting from a point a going to infinity?



#7
Sep911, 10:39 AM

P: 62

i imagined that it may be a straight line i.e. the shortest distance between infinity and P



#8
Sep911, 12:57 PM

P: 3,014

This straight line is called a ray, and the direction is called radial direction. On it, [itex]ds = dr[/itex]. You need to use this.



#9
Sep1011, 12:17 AM

P: 62

I am confused with the signs and directions of ds ,dr and test charge and also
how work done by external element is negative of work done by electric field. 


#10
Sep1011, 03:27 AM

Sci Advisor
Thanks
P: 2,302

I never understood, why integration in tensor calculus is obscured by some awkward notation. I guess, many textbook writers think, it's more intuitive to work with angles instead of vectors, but that's not true. To calculate a line integral, it's much more convenient to use the definition of that type of integral. Let [itex]\vec{V}(\vec{x})[/itex] be a vector field, defined in some domain of [itex]\mathbb{R}^3[/itex] and [itex]C: \lambda \in \mathbb{R} \supseteq (a,b) \mapsto \vec{x}(\lambda) \in \mathbb{R}^3[/itex] with values in the definition domain of [itex]\vec{V}[/itex]. Then the line integral over the vector field along the path [itex]C[/itex] is defined as
[tex]\int_C \mathrm{d} \vec{x} \cdot \vec{V}(\vec{x})=\int_{a}^{b} \mathrm{d} \lambda \frac{\mathrm{d} \vec{x}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{V}[\vec{x}(\lambda)].[/tex] 


#11
Sep1211, 10:04 AM

P: 17

The lines of forces are very simple in this case.
Now you don't have to worry about the signs and angles here. Just simply use the formula for moving a charge through an electric field and you'll get it. I hope this helps. Thank you. 


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