To calculate potential due to continuous charge distribution at a point

Step 1

• Choose element such that varying the element gives us whole charge distribution.

Step 2

• Calculate charge on that element.

Step 3

• Calculate potential due to that element at a point.

Step 4

• Total potential is the scalar sum of potentials of all the elements.

Step 5

\(V= \int \dfrac {1 dq}{4\pi\epsilon_0r}\)

Case-I  Thin Rod

• Consider a thin rod of length L  and charge + Q which is equally distributed over its length.
• To calculate potential at point P, take a small element of length dx which acts as a point charge.
• Charge on the element,

           \(dQ=\dfrac {Q}{L}dx\) [ charge of element = charge per unit length × length of element ]

Case-II  Ring

• Consider a thin ring of radius R and charge + Q. To calculate potential at distance x on its axis from center, choose a small element of length \(d\ell\), as shown in figure.
• \(d\ell\) element acts as a point charge.
• Charge on the element,

          \(dQ=\dfrac {Q}{2\pi R}d\ell\) [ Charge of element = charge per unit length × length of element ]

Case-III   Disk 

• Consider a disk of radius R and charge + Q. To calculate potential at its axial point, take a small ring as an element.
• Total potential is the scalar sum of potentials due to all the elements.
• Charge on this small ring, \(dq=\dfrac {Q}{\pi R^2}(2\pi rdr)\)

[Charge on element = Charge per unit area × Area of element]

Case-I   Ring

• Consider a ring of radius R and charge +Q distributed over its length.
• To find potential at its center O, consider an element of charge dq at the circumference of ring.
• Electric potential at the center of ring due to small element of charge dq,

           \(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)

• The element acts as a point charge. As all elements are at distance R from this point. So, potential due to all small elements will be same.
• Total potential at the center will be scalar sum of potentials due to all elements. So, total potential at center O,

          \(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)

          \(V=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q}{R}\)         \(\left[\because \;\int dq=+Q \;\right]\)

Case-II   Hollow sphere

• Consider a hollow sphere of radius R with charge +Q distributed over its surface.
• To calculate potential at its center O, choose a small element of charge dq on its surface.
• This element will act as a point charge.
• Electric potential at the center due to this element 

             \(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)

• Since all the elements on the surface are at equal distance from center,so potential due to all the elements will be same.
• Total potential at the center will be scalar sum of potentials of all the elements on the surface.
• Total electric potential at the center O,

            \(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{R}\)

            \(V=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q}{R}\)         \(\left[\because \;\int dq=+Q \;\right]\)

• Consider a disk of radius  R  with charge  +Q distributed over its surface.

• To calculate potential at a point P at a distance  x on its axis, use the method of elements.
• Consider a small ring as an element at a distance r from center and width  dr
• Charge on an element

\(dq=\dfrac {Q}{\pi R^2}(2\pi\; r\;dr)\)

(Charge on an element = Charge per unit area × Area of element )

• Potential at point P due to this element,

\(dV=\dfrac {dq}{4\pi\epsilon_0\sqrt {r^2+x^2}}\)

Since \(dq=\dfrac {Q}{\pi R^2}(2\pi\; r\;dr)\)

\(dV=\dfrac {Q\; (2\pi\; r\;dr)}{\pi R^2\times 4\pi\epsilon_0\sqrt {r^2+x^2}}\)

\(dV=\dfrac {Q.\; r\;dr}{2\pi\epsilon_0 R^2\sqrt {r^2+x^2}}\)

• ?Integrating the potential from  r = 0  to  r = R  gives the total potential of whole disk,

\(V=\int\limits_0^R\dfrac {Q}{2\pi\epsilon_0R^2} \dfrac {rdr}{\sqrt{r^2+x^2}}\)

\(V=\dfrac {Q}{2\pi\epsilon_0R^2} \int\limits_0^R \dfrac {rdr}{\sqrt{r^2+x^2}}\)

\(Put\,\,\,\,{r^2 + x^2}={t^2}\)

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, \(V=\dfrac {Q}{2\pi\epsilon_0R^2} \int\limits_x^\sqrt{R^2+x^2} \dfrac {t\; dt}{t}\)}

^{\(V=\dfrac {Q}{2\pi\epsilon_0R^2} [\;t\;]_x^\sqrt{R^2+x^2} \)}

^{\(V=\dfrac {Q}{2\pi\epsilon_0R^2} [\sqrt{R^2+x^2}-x]\)}

Steps to calculate potential at any point due to combinations

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.

• Ring can be considered as combination of point charges and total potential at point P is the scalar sum of potentials due to all charges.
• Total potential at point P,
• \(V=\int\dfrac {dq}{4\pi\epsilon_0\sqrt {R^2+x^2}}\)

  \(V=\dfrac {q}{4\pi\epsilon_0\sqrt {R^2+x^2}}\)

   

• Consider a thin rod of length L and charge +Q .
• To calculate electric potential at a point P on its axis situated at a distance r from the center of the rod, use method of element.
• Choose an element at a distance  x  from center of rod of length dx.
• This element acts as point charge.
• Charge on this element,

           \(dq=\dfrac {Q}{L}\times dx\) [ Charge on element = charge per unit length × length of the element]

• To calculate electric potential at point P due to this element, first calculate its distance from point P. This distance will be (r–x).
• Electric potential at point P due to this element, 

           \(dV=\dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{(r-x)}\)

• Electric potential is a scalar quantity.
• Total potential at point P is the scalar sum of potentials due to all small elements.
• Total electric potential 

\(V=\int \dfrac {1}{4\pi\epsilon_0}\dfrac {dq}{(r-x)}\)

\(V=\int\limits_{-L/2}^{L/2} \dfrac {1}{4\pi\epsilon_0} \;\dfrac {Qdx}{L(r-x)}\)

\(V=\dfrac {1}{4\pi\epsilon_0} \;\dfrac {Q}{L} \int\limits_{-L/2}^{L/2} \;\dfrac {dx}{r-x}\)

\(V=\dfrac {1}{4\pi\epsilon_0} \;\dfrac {Q}{L} [-\ell n(r-x)]_{-L/2}^{L/2}\)

\(\because \) \(\int\limits_m^n \left( \;\dfrac {1}{a-x} \right) dx=[-\ell n(a-x)]_{m}^{n}\)

\(V= \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{L} \left[ -\ell n \left (r-\dfrac {L}{2}\right)+\ell n \left (r+\dfrac {L}{2}\right) \right]\)

\(V= \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{L} \ell n \left( \dfrac {r+L/2}{r-L/2} \right)\)