**let us discuss and derive the electric potential due to single charge or point charge. **We will calculate electric potential at any point P due to a single point charge +q at O ;where OP=r

Electric potential at P is the amount of work done in carrying a unit positive charge from ∞ to P.

At any point A on the line joining OP ,where OA=x,the electric intensity is E=1/4πε_{0}q/x^{2} along OA produced (try to make the figure yourself).

Amount of work done in moving a unit positive charge from A to B where (AB=dX) is

dW=E.dx=Edx cos 180^{0}

= -Edx

Total work done in moving unit + charge from ∞ to the point P is

W= -∫Edx

=∫ -1/4πε_{0}q/x^{2}dx

= -q/4πε_{0}∫x^{-2} dx

= -q/4πε_{0}[-1/x]

W= q/4πε_{0}[1/r-1/∞]

=q/4πε_{0}r

This is the potential at P i.e

V=W=q/4πε_{0}r

When q is positive ,potential V is positive and when q is negative,potential V is negative i.e a positive charge produces a positive electric potential and negative charge produces a negative electric potential.

This is the electric potential due to single charge or point charge.

*Related*

This entry was posted in Uncategorized. Bookmark the

permalink.