Question

Q. Derive an Expression for electric potential due to a point charge ​

Answer 1

Answer:

Explanation:

Consider the electric potential due to a point charge q, As we move from point A, at distance rA from the charge q, to point B, at distance rB from the charge q, the change in electric potential is

ΔVBA​=VB​−VA​=−∫AB​E.ds

E.ds=[kr2q​]r^.ds

r^.ds=dr

Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.

ΔVBA​=VB​−VA​=−∫rA​rB​​Edr=−∫rA​rB​​[kr2q​dr]

ΔVBA​=VB​−VA​=−kq[(−1)r−1]rA​rB​​

ΔVBA​=VB​−VA​=kq[rB​1​−rB​1​]

This is the change in electric potential due to a point charge as we move from rA to rB.

We could ask about the change in electric potential energy as we move a charge q' from radius rA to rB due to a point charge q

ΔUBA​=kq′q[rB​1​−rA​1​]

As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,

rA​=∞

That means we can then write the electric potential at some radius r as V=kqr1​

Answer 2

$\bold \color{green} { \underline { \underline \red{ \blue \bigstar \: required \: answer \blue \star}}}$

Consider the electric potential due to a point charge q, As we move from point A, at distance rA from the charge q, to point B, at distance rB from the charge q, the change in electric potential is

$\triangle \: VBA = VB - VA = −∫ \binom{B}{A}E.ds$

$\implies$

$\bold {E.ds =( {K \frac{q}{r²})r^.ds. r^.ds = dr}}$

Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.

$\triangle \: VBA = VB - VA = - ∫ \binom{rB}{rA} (K \frac{q}{ {r}^{2} } .dr)$

$\triangle \: VBA = VB - VA = - Kq( - 1r - 1) \binom{rB}{rA}$

$\triangle \: VBA = VB - VA \: = Kq( \frac{1}{rB1} - \frac{1}{rB} )$

This is the change in electric potential due to a point charge as we move from rA to rB.

We could ask about the change in electric potential energy as we move a charge q' from radius rA to rB due to a point charge q.

$\triangle \: UBA = Kq′q( \frac{1}{rB} - \frac{1}{rA} )$

As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,

$rA = \infty$

That means we can then write the electric potential at some radius r as

$V \: = Kq \frac{1}{r}$