Question

derive an expression for the electric potential due to point charge​

Answer 1

Defination :-

Electric potential at a point may be defined as Work done in bringing a unit positive test charge from infinity to that point without changing kinetic energy.

⇝ In Attached Figure :-

• q = Point charge due to which electric potential is to be find.

• $\sf+ q_ \circ$ = +ve Test charge

• P = Any point where potential is to be find.

⇝ Derivation :-

[ Figure in Attachment ]

Small work done to move charge from point P to infinity is :

$\text{dW = - F.dx}$

$:\longmapsto\text{dW} = - \frac{\text{kqq}_{\circ}}{\text x {}^{2} }$

⏩ Integrating Both Side ;

$\text{W}_{\text P\rightarrow \infty } = - \int\limits^ \infty _\text r \text{kqq}_{\circ}( {\text x}^{ - 2} ) \ \text {dx}$

$= - \text{kqq}_{\circ} {{\bigg[ - \frac{1}{\text x} \bigg]}_\text r^ \infty }$

$\text{W}_{\text P \rightarrow \infty } = - \frac{\text{kqq}_{\circ}}{\text r}$

$\large\red{\therefore \: \boxed{ \boxed{\text{W}_{\infty \rightarrow\text P } = \frac{\text{kqq}_{\circ}}{\text r} }}}$

⏩ By Defination :

$:\longmapsto \rm V_P=\frac{ W_{\infty \rightarrow P} }{q_{\circ}}$

$:\longmapsto \rm V_P = \frac{{kqq}_{\circ}}{rq_{\circ}}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf V_P = \frac{{kq}}{r}} }}}$

which is electric potential due to a point charge at point P.