Derivation of electric potential due to point charge

[Sumedh]

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:
V=-\int \vec{E}\cdot d\vec{s}
V=-\int E\cdot ds cos \vartheta
if the stationary charge is positive and
if the test charge is is moved from infinity to point P
then
V=-\int E\cdot ds cos 180
V=-KQ \int \frac {1}{r^2} ds cos 180

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.

 

[Dickfore]

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

 

[Sumedh]

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

 

[Dickfore]

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

 

[Sumedh]

it should have relation with
\frac {1}{r^2}

 

[Dickfore]

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

 

[Sumedh]

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

 

[Dickfore]

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

 

[Sumedh]

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.

 

[vanhees71]

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 \vec{V}(\vec{x}) be a vector field, defined in some domain of \mathbb{R}^3 and C: \lambda \in \mathbb{R} \supseteq (a,b) \mapsto \vec{x}(\lambda) \in \mathbb{R}^3 with values in the definition domain of \vec{V}. Then the line integral over the vector field along the path C is defined as

\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)].

 

[Bhargava2011]

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.