Practice to calculate potential due to continuous charge distribution at a point and circular ring and hollow sphere at the center, at the axis of a ring, and at the axis of a disk.
Step 1
Choose element such that varying the element gives us whole charge distribution.
Step 2
Step 3
Calculate potential due to that element at a point.
Step 4
Step 5
\(V= \int \dfrac {1 dq}{4\pi\epsilon_0r}\)
\(dQ=\dfrac {Q}{L}dx\) [ charge of element = charge per unit length × length of element ]
\(dQ=\dfrac {Q}{2\pi R}d\ell\) [ Charge of element = charge per unit length × length of element ]
[Charge on element = Charge per unit area × Area of element]
A A disk of radius r = 5 cm and charge of element = \(10^3r^2\,C\)
B A ring of radius \(r\,'\) , width dr and charge of element = \(4\times10^3\,r'\,dr\)
C A ring of radius \(r\,'\), width dr and charge of element = \(2\times10^3r^2\,C\)
D A rod of length \(\ell\) at distance x from center , width dx and charge of element = \(2\times10^3\times x\; dx \,C\)
\(V=\dfrac {q}{4\pi\epsilon_0\sqrt {R^2+x^2}}\)
\(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)
\(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)
\(V=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q}{R}\) \(\left[\because \;\int dq=+Q \;\right]\)
\(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)
\(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)
\(V=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q}{R}\) \(\left[\because \;\int dq=+Q \;\right]\)
\(dq=\dfrac {Q}{L}\times dx\) [ Charge on element = charge per unit length × length of the element]
\(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{(r-x)}\)
\(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{(r-x)}\)
\(V=\int\limits_{-L/2}^{L/2} \dfrac {1}{4\pi\epsilon_0} \;\dfrac {Qdx}{L(r-x)}\)
\(V=\dfrac {1}{4\pi\epsilon_0} \;\dfrac {Q}{L} \int\limits_{-L/2}^{L/2} \;\dfrac {dx}{r-x}\)
\(V=\dfrac {1}{4\pi\epsilon_0} \;\dfrac {Q}{L} [-\ell n(r-x)]_{-L/2}^{L/2}\)
\(\because \) \(\int\limits_m^n \left( \;\dfrac {1}{a-x} \right) dx=[-\ell n(a-x)]_{m}^{n}\)
\(V= \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{L} \left[ -\ell n \left (r-\dfrac {L}{2}\right)+\ell n \left (r+\dfrac {L}{2}\right) \right]\)
\(V= \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{L} \ell n \left( \dfrac {r+L/2}{r-L/2} \right)\)
\(dq=\dfrac {Q}{\pi R^2}(2\pi\; r\;dr)\)
(Charge on an element = Charge per unit area × Area of element )
\(dV=\dfrac {dq}{4\pi\epsilon_0\sqrt {r^2+x^2}}\)
Since \(dq=\dfrac {Q}{\pi R^2}(2\pi\; r\;dr)\)
\(dV=\dfrac {Q\; (2\pi\; r\;dr)}{\pi R^2\times 4\pi\epsilon_0\sqrt {r^2+x^2}}\)
\(dV=\dfrac {Q.\; r\;dr}{2\pi\epsilon_0 R^2\sqrt {r^2+x^2}}\)
\(V=\int\limits_0^R\dfrac {Q}{2\pi\epsilon_0R^2} \dfrac {rdr}{\sqrt{r^2+x^2}}\)
\(V=\dfrac {Q}{2\pi\epsilon_0R^2} \int\limits_0^R \dfrac {rdr}{\sqrt{r^2+x^2}}\)
\(Put\,\,\,\,{r^2 + x^2}={t^2}\)
By partial differentiation, \({ 2r\,+0}=2t\,dt\)
\(\rightarrow r\,dr=t\,dt\)
\(Limits\rightarrow\,\,\,\,{r^2 + x^2}={t^2}\)
\(at\,r=0\,\,\,\,\,\,{t^2}={x^2} \\ \,\,\,\,\,\,\,\,\,\,\,t=x \\at\,r=R\,\,\,\,\,{R^2+x^2}={t^2} \\\,\,\,\,\,\,\,\,\,\,t=\sqrt{R^2+x^2}\)
^{So, \(V=\dfrac {Q}{2\pi\epsilon_0R^2} \int\limits_x^\sqrt{R^2+x^2} \dfrac {t\; dt}{t}\)}
^{\(V=\dfrac {Q}{2\pi\epsilon_0R^2} [\;t\;]_x^\sqrt{R^2+x^2} \)}
^{\(V=\dfrac {Q}{2\pi\epsilon_0R^2} [\sqrt{R^2+x^2}-x]\)}
Step 1
Calculate potential due to different-different charge distribution.
Step 2
Total potential is the scalar sum of potentials due to all charge distributions.
A 3.5 × 1012 V
B 9.5 × 1015 V
C 9.2 × 108 V
D 1.1 × 106 V