12.5 Potential Due to Uniformly charged Disc

  Electrostatics

<p>$${V_P} = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{q}{{PB}} - \frac{q}{{PA}}} \right]$$ When&nbsp;$$r \gg a,\;PB = r - a\;\cos \;\theta $$&nbsp;$$PA = r + a\;\cos \;\theta $$&nbsp;$${V_p} = \frac{1}{{4\pi { \in _0}}}.\frac{{2qa\;\cos \theta }}{{\left( {{r^2} - {a^2}{{\cos }^2}\theta } \right)}}$$&nbsp;</p><p style="text-align: center; "><span style="background-color: rgb(255, 255, 0);">Diagram</span></p><p>As $r \gg a,{a^2}\;{\cos ^2}\theta $ can be neglected in comparison to ${{r^2}}$. $$\therefore \quad {V_p} = \frac{1}{{4\pi { \in _0}}}.\frac{{2\;q\;a\;\cos \theta }}{{{r^2}}} = \frac{{p\;\cos \;\theta }}{{4\pi {\varepsilon _o}{r^2}}}$$</p>