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Potential Due To Continuous Charge Distribution

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.

Potential due to Continuous Charge Distribution

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]

Illustration Questions

The value of electric potential due to a disk of charge Q = +5C and radius r = 5 cm is to be calculated at an axial point(P). Choose the correct choice of element and charge on 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 = \(2\times10^3\times x\; dx \,C\)

×

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

image

Charge of element = charge per unit area × area of element

Charge per unit area = \(\dfrac {5}{\pi(5\times10^{-2})^2}=\dfrac {5}{\pi\times 25\times10^{-4}}\)

image

Area of element =  \(2\pi\,r\,'dr\)

Charge of element  =  \(\dfrac {5}{\pi\times 25\times10^{-4}}\times 2\pi r\,'dr\)

                             \(=4\times10^3\, r\,' dr \,C\) 

The value of electric potential due to a disk of charge Q = +5C and radius r = 5 cm is to be calculated at an axial point(P). Choose the correct choice of element and charge on element.

A

A disk of radius r = 5 cm and charge of element = \(10^3r^2\,C\)

.

image
B

A ring of radius \(r\,'\) , width dr and charge of element =  \(4\times10^3\,r'\,dr\)

image
C

A ring of radius \(r\,'\), width dr and charge of element = \(2\times10^3r^2\,C\)

image
D

A rod of length \(\ell\) at distance x from center , width dx and charge of element = \(2\times10^3\times x\; dx \,C\)

image

Option B is Correct

Potential at the Axis of a Ring

  • 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,
  • \(V=\int\dfrac {dq}{4\pi\epsilon_0\sqrt {R^2+x^2}}\)

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

     

Illustration Questions

Choose the correct option of steps for obtaining potential due to a ring at a point on its axis.

A

B

C

D

×

To calculate potential due to continuous charge distribution at a point.

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

Calculate charge on that element.

Calculate potential due to that element at a point.

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

 

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

Choose the correct option of steps for obtaining potential due to a ring at a point on its axis.

A image
B image
C image
D

Option B is Correct

Potential due to Circular Ring and Hollow Sphere at the Center

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,

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

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

 

Illustration Questions

A hollow sphere of radius \(R=+5\,cm\) is given a charge \(Q=5\,\mu\,C\). Calculate the electric potential at its center. \(\left [ \dfrac {1}{4\pi\epsilon_0} =9\times 10^9\;Nm^2/C^2 \right]\)

A 0 V

B 1 × 105 V

C 2 × 105 V

D 9 × 105 V

×

Electric potential of hollow sphere of radius R and charge +Q at its center = \(\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}\)

  \(Q=5\,\mu\,C\) = \(5\times10^{-6}\;C\) and \(R=5\,cm\) = \(5\times10^{-2}\;m\)

\(V=9\times10^9\times\dfrac {5\times10^{-6}}{5\times 10^{-2}} \\=9\times10^5\;V\)

A hollow sphere of radius \(R=+5\,cm\) is given a charge \(Q=5\,\mu\,C\). Calculate the electric potential at its center. \(\left [ \dfrac {1}{4\pi\epsilon_0} =9\times 10^9\;Nm^2/C^2 \right]\)

A

0 V

.

B

1 × 105 V

C

2 × 105 V

D

9 × 105 V

Option D is Correct

Potential at the Axis of  a Charged Rod

  • 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 

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

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

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

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

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

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

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

Illustration Questions

Choose the correct sequence of steps to calculate electric potential due to charged rod at its axis.

A

B

C

D

×

To calculate potential due to continuous charge distribution at a point.

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

Calculate charge on that element.

Calculate potential due to that element at a point.

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

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

Choose the correct sequence of steps to calculate electric potential due to charged rod at its axis.

A image
B image
C image
D

Option C is Correct

Potential at the Axis of a Disk

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

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

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

Illustration Questions

Choose the option for correct sequence of steps for obtaining potential due to a ring at a point on its axis.

A

B

C

D

×

To calculate potential due to continuous charge distribution at a point.

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

Calculate charge on that element.

Calculate potential due to that element at a point.

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

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

Choose the option for correct sequence of steps for obtaining potential due to a ring at a point on its axis.

A image
B image
C image
D

Option B is Correct

Potential at any point due to combinations

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.

Illustration Questions

Calculate electric potential at point  \(P\) at a distance \(x=4\,cm\) from the center on the common axis of two concentric circular objects. The inner one is a disk of radius \(R_D=3\,cm\)  and charge \(Q_D=-5\,C\) and outer one is the ring of radius \(R_R=6\,cm\)  and charge \(Q_R=10\sqrt{13}\,C\) .\(\left [ \dfrac {1}{4\pi\epsilon_0}=9\times10^9\; Nm^2/C^2 \right]\)

A 3.5 × 1012 V

B 9.5 × 1015 V

C 9.2 × 108 V

D 1.1 × 106 V

×

Potential at point P due to the disk,

x = 4 cm, RD = 3 cm and QD = –5 C

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

\(V_{Disk}= \dfrac {(-5)\times2\times 9\times10^9}{(3\times10^{-2})^2} [\sqrt{(3\times10^{-2})^2+(4\times10^{-2})^2}-(4\times10^{-2})]\)

\(V_{Disk}=-10^{12}\; Volt\)

image

Potential at point P due to the ring,

\(x=4\,cm\) , \(R_R=6\,cm\)  and \(Q_R=10\sqrt{13}\,C\)

\(V_{Ring}= \dfrac {1}{4\pi\epsilon_0} \times \dfrac {Q}{\sqrt{R^2+x^2}} \)

\(V_{Ring}= \dfrac {9\times10^9\times 10\sqrt{13}}{\sqrt {(6)^2+(4)^2}} \)

\(V_{Ring}= \dfrac {9\times10^9\times 10\sqrt{13}}{2\sqrt {13}\times10^{-2}} \)

  \(V_{Ring}=45\times10^{11}\)                                                    

\(V_{Ring}=4.5\times10^{12}\; Volt\)

image

Total electric potential at point P,

\(V=V_{Disk}+V_{Ring}\)

\(V=(-10^{12})+4.5\times10^{12}\)

\(V=3.5\times10^{12}\; Volt\)

Calculate electric potential at point  \(P\) at a distance \(x=4\,cm\) from the center on the common axis of two concentric circular objects. The inner one is a disk of radius \(R_D=3\,cm\)  and charge \(Q_D=-5\,C\) and outer one is the ring of radius \(R_R=6\,cm\)  and charge \(Q_R=10\sqrt{13}\,C\) .\(\left [ \dfrac {1}{4\pi\epsilon_0}=9\times10^9\; Nm^2/C^2 \right]\)

image
A

3.5 × 1012 V

.

B

9.5 × 1015 V

C

9.2 × 108 V

D

1.1 × 106 V

Option A is Correct

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