Informative line

Electric Field Due To Continuous Charge Distribution

Learn calculation of electric field due to uniformly charged ring and disk along the central axis, practice electric field continuous charge distribution.

Selection of Element and Calculation of Charge on Element

  • Selection of element of a continuous charge distribution should be done such that summing up all the elements gives the entire charge distribution.
  • Total charge on continuous charge distribution can be evaluated as the summation of charge of all the charge elements (that is, by applying the superposition principle).

 

Case 1 : Thin Rod

  • Consider a rod of length L having linear charge density  \(\lambda\) and total charge Q.
  • Consider an element of rod of length dx at a distance x from one end, as shown in figure. 
  • Selection of element is done such that summing up all the elements gives the entire charge distribution.

Charge on rod of length L = Q

Charge on rod of unit length = \(\dfrac {Q}{L}\)

Charge on element of rod of length dx = \(\dfrac {Q}{L}×dx\), where, \(\dfrac {Q}{L}=\lambda\)

Charge on element \(dQ=\lambda\; dx\)

NOTE: Linear charge density is defined as the charge per unit length when charge is distributed uniformly over the length.

 

Case 2 : Thin Ring

  • Consider a ring of radius R having a uniformly distributed positive charge Q.
  • Consider a small element of length dx and having charge dQ on it.
  • Selection of element should be done such that summing up all the elements gives the whole ring (charge distribution).

Charge on the small element  \( dQ = dx ×\text{ charge density}\)

\(dQ=dx×\dfrac {Q}{2\pi R}\)

 

Case 3 : Disk

  • Consider a disk of radius R, having a uniform charge density \(\sigma\) and a total charge Q uniformly distributed over it.
  • Choose an element of disk of radius x and thickness dx having charge dQ.
  • Selection of element should be done such that summing up all the elements gives the whole disk.

Charge on elemental ring dQ = Area of element × Surface charge density

\(dQ=2\;\pi\,x\,dx×\dfrac {Q}{\pi\,R^2}\)

NOTE: Surface charge density (\(\sigma\)) is defined as the charge per unit area (charge Q is uniformly distributed over area).

Illustration Questions

A charge Q = 5 C is uniformly distributed over the whole length L = 5 m of rod. Choose the correct option for the selection of element and charge on this element.

A Charge on element dQ = dx

B Charge on element dQ = dx

C Charge on element dQ = L dx

D Charge on element dQ = L dx

×

Selection of element is done such that summing up all the elements gives the entire charge distribution.

image

Charge on element = length of element × Linear charge density (\(\lambda\))

\(dQ=dx×\dfrac {Q}{L}\)

\(dQ=dx×\dfrac {5C}{5m}\)

\(dQ = dx\,\, \)

image

Hence, option (B) is correct.

image

A charge Q = 5 C is uniformly distributed over the whole length L = 5 m of rod. Choose the correct option for the selection of element and charge on this element.

A

Charge on element dQ = dx

.

image
B

Charge on element dQ = dx

image
C

Charge on element dQ = L dx

image
D

Charge on element dQ = L dx

image

Option B is Correct

Calculation of Electric Field due to Uniformly Charged Ring

  • Consider a ring of radius R, having charge Q uniformly distributed over it.

  • To calculate electric field at point (P) on the axis of ring placed in Y-Z plane, choose the elements on the ring.

  • The net electric field at point P can be calculated by taking components of electric field due to all elements.
  • Perpendicular component of electric field at P due to one element is cancelled by other element. Hence, net electric field at point P will be only in x-direction due to all elements.

\(\vec {dE}=\dfrac {K\cdot dQ}{r^2}\)

\(\vec {dE}=\dfrac {K\cdot dQ}{\sqrt {R^2+a^2}}\)

So, Electric field in x-direction

\(dE_x=\dfrac {K\cdot dQ}{R^2+a^2}\,cos\,\theta\)

  • The total electric field in x-direction due to ring is given as \(\vec E=\int dE_x\; \hat i\)

\(\vec E=\int \dfrac {K\cdot dQ}{R^2+a^2}cos\;\theta\; \hat i\)

 

 

For cos \(\theta\) :

\(cos\,\theta=\dfrac {a}{\sqrt {R^2+a^2}}\)

\(\vec E=\int dE_x\,\hat i=\int \dfrac {K.dQ}{(R^2+a^2)}\cdot\dfrac {a}{\sqrt {R^2+a^2}}\,\hat i\)

\(\vec E=E_x\,\hat i=\dfrac {Ka\int dQ}{(R^2+a^2)^{3/2}}\,\hat i\)

\(\vec E=\dfrac {K\,Q\,a}{(R^2+a^2)^{3/2}}\,\hat i\)

  • In general, the electric field due to uniformly charged (Q) ring of radius R at a distance 'a' from the center on the axis of ring is given as: 

\(\vec E=\dfrac {K\,Q\,a}{(R^2+a^2)^{3/2}}\,\hat r\)

where, \(\hat r\) is the unit vector along the axis of ring.

Illustration Questions

A charge of Q = 125 nC is uniformly distributed over a ring of radius R = 3m, placed in Y-Z plane. Calculate the electric field at a distance a = 4 m from the center on the axis of ring.

A 12 N/C

B 24 N/C

C 36 N/C

D 22 N/C

×

The electric field due to uniformly charged (Q) ring of radius R at a distance 'a' from the center on the axis of ring is given as :

\(\vec E=\dfrac {K\,Q\,a}{(R^2+a^2)^{3/2}}\,\hat r\)

where, \(\hat r\) is the unit vector along the axis of ring

image

Since unit vector is along x-axis, i.e. in x-direction

So,  \(\hat r = \hat i\)

image

Given :  Q = 125 nC, R = 3m, a = 4m

\(\vec E=\dfrac {9×10^9×125×10^{-9}×4}{(4^2+3^2)^{3/2}}\,\hat i\)

\(\vec E=\) 36 N/C

image

A charge of Q = 125 nC is uniformly distributed over a ring of radius R = 3m, placed in Y-Z plane. Calculate the electric field at a distance a = 4 m from the center on the axis of ring.

A

12 N/C

.

B

24 N/C

C

36 N/C

D

22 N/C

Option C is Correct

Calculation of Electric Field due to Uniformly Charged Disk

  • Consider a disk of radius R having a uniform surface charge density \(\sigma\).

  • To calculate electric field at point P which is at x distance away the center along the axis of ring, choose the differential element.
  • Selection of element should be done in such a way that summing up all the elements gives the whole disk.
  • The differential element will be a concentric ring of radius 'r' and thickness 'dr' having charge 'dq' on it.

\(dq =\sigma\,dA\)

\(dq =\sigma(2\pi r\,dr)\)

\(dq =2\pi\,\sigma\, r\,dr\)

  • Taking differential element separately from disk and calculating electric field due to this small elemental ring having uniformly distributed charge 'dq'.

  • Electric field due to this small elemental ring

 \(d\vec E=\dfrac {K\, dq}{r^2+x^2}\)

  • Since, the Y-component gets canceled and the electric field will be only due to contribution of X-component in X-direction.

\(d\vec E_x=\dfrac {K}{(r^2+x^2)}\,cos\,\theta (2\pi r\,dr\,\sigma)\)

\(d\vec E_x=\dfrac {K}{(r^2+x^2)}×\dfrac {x}{\sqrt {x^2+r^2}}\,(2\pi\,\sigma\, r\,dr\,)\)       \(\left [ \because cos\theta=\dfrac {x}{\sqrt {x^2+r^2}} \right]\)

\(d\vec E_x=\dfrac {K\,x\,(2\pi\,\sigma\, r\,dr\,)}{(r^2+x^2)^{3/2}}\)

Integrating both sides,

\(\int d\vec E_x=K\,x\,(2\pi\, \sigma)\displaystyle\int\limits_0^R\dfrac{r\,dr}{(r^2+x^2)^{3/2}}\)

\(\vec E=2\pi\, K\,x\,\sigma\;\displaystyle\int\limits_0^R\dfrac{r\,dr}{(r^2+x^2)^{3/2}}\hat i\)

Let \(r^2+x^2=t\) then \(2r\,\dfrac{dr}{dt}=1\)

\(\dfrac{dt}{2}=r\,dr\)

So, \(\vec E=2\,\pi Kx\sigma \displaystyle\int\limits_0^R\dfrac{dt}{2t^{3/2}}\hat i\)

\(\vec E=\dfrac{2\,\pi Kx\sigma}{2} \displaystyle\int\limits_0^R{t^{-3/2}dt}\,\hat i\)

\(\vec E=\pi\,Kx\sigma \Bigg[\dfrac{t^{-3/2+1}}{{-3/2}+1}\Bigg]_0^R\,\hat i\)

\(\vec E=\pi Kx\sigma \times \Bigg[\dfrac{-2}{\sqrt t}\Bigg]_0^R\,\,\hat i\)

\(\vec E=2\pi K x\sigma \Bigg[\dfrac{-1}{\sqrt{r^2+x^2}}\Bigg]_0^R \,\hat i\)

\(\vec E=2\pi Kx\,\sigma\Bigg[\dfrac{-1}{\sqrt{R^2+x^2}}+\dfrac{1}{x}\Bigg] \,\hat i\\ \vec E=2\pi K\sigma\Bigg[1-\dfrac{x}{\sqrt{R^2+x^2}}\Bigg]\,\hat i\)

 

  • In general,

\(\vec E=2\pi\, K\,\sigma\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)

 \(=\dfrac {2\pi\,\sigma}{4\,\pi\epsilon_0}\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)

\(=\dfrac {\sigma}{2\,\epsilon_0}\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)

where, \(\hat r\)is the unit vector along axis.

Illustration Questions

Calculate the electric field at a distance x = 4m along the axis from the center of disk of radius R = 3m, having surface charge density  \(\sigma=\)5 nC/m2 placed in Y-Z plane.

A 16 \(\pi\,\hat i\) N/C

B 18 \(\pi\,\hat i\) N/C

C 8 \(\pi\,\hat i\) N/C

D 3 \(\pi\,\hat i\) N/C

×

Electric field due to a disk of radius R having surface charge density \(\sigma\) at distance x along axis is given as 

\(\vec E=2\,\pi\,K\,\sigma \left [ 1-\dfrac {x}{\sqrt {R^2+x^2}} \right ]\,\hat r\)

where, \(\hat r\)is the unit vector along axis.

image

Since, the disk is in Y-Z plane so, its unit vector will be in x-direction 

\(\hat r=\hat i\)

image

Given : x = 4m, R = 3m,  \(\sigma=\)5 nC/m2

\(\vec E=2\pi×9×10^9×5×10^{-9} \left [ 1-\dfrac {4}{\sqrt{ (4)^2+(3)^2}} \right ]\,\hat i\)

\(\vec E=18\pi\,\hat i \,\text{N/C}\)

image

Calculate the electric field at a distance x = 4m along the axis from the center of disk of radius R = 3m, having surface charge density  \(\sigma=\)5 nC/m2 placed in Y-Z plane.

A

16 \(\pi\,\hat i\) N/C

.

B

18 \(\pi\,\hat i\) N/C

C

\(\pi\,\hat i\) N/C

D

\(\pi\,\hat i\) N/C

Option B is Correct

Electric Field due to Section of Ring at its Center

  • Consider a section of ring of radius R, as shown in figure.

  • To calculate electric field at the center of ring due to section of ring, choose the element on the ring.
  • An element of length \(d\ell\) subtending an angle \(d\alpha\) at the center of the ring, as shown in figure.
  • The horizontal components of electric field will cancel out and vertical component will get added, as shown in figure.
  • The net electric field at point O

\(dE_{net}=2d\,E cos\,\alpha\)

Integrating both sides,

\(\int dE_{net}=\int\limits_0^{\theta/2}\,2d\;E\;cos\alpha\)

\(E_{net}=2\int\limits_0^{\theta/2}\,\dfrac {\lambda\;d\ell}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

From figure,

\(d\ell=R\,d\alpha\)

\(\vec E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;R\,d\alpha}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R}\Big[sin\;\alpha\Big]_0^{\theta/2}\)

\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R} \Big[sin\;\dfrac {\theta}{2}-sin\;0\Big]\)

\(\vec E_{net}=\dfrac {\lambda}{2\,\pi\,\epsilon_0R}sin\;\dfrac {\theta}{2}\)

 

Illustration Questions

Calculate the electric field due to a semicircular portion of ring radius R = 3m having linear charge density  \(\lambda=\)1 nC/m at its center.

A 20 N/C

B 8 N/C

C 5 N/C

D 6 N/C

×

An element of length \(d\ell\) subtending an angle \(d\alpha\) at the center of the ring, as shown in figure.

The horizontal components of electric field will cancel out and vertical component will get added, as shown in figure.

image image

The horizontal components of electric field will cancel out and vertical component will get added, as shown in figure.

 

image

The net electric field at point P

\(dE_{net}=2\,dE\, cos\,\alpha\)

Integrating both sides,

\(\int dE_{net}=\int\limits_0^{\theta/2}\,2\,dE\;cos\alpha\)

\(E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;d\ell}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

 

image

From figure,

\(d\ell=R\,d\alpha\)

From figure, angle varies from 0 to \(\dfrac {\theta}{2}\) for one element.

\(\vec E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;R\,d\alpha}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R}\Big[sin\;\alpha\Big]_0^{\theta/2}\)

\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R} \Big[sin\;\dfrac {\theta}{2}-sin\;0\Big]\)

\(\vec E_{net}=\dfrac {\lambda}{2\,\pi\,\epsilon_0R} sin\;\dfrac {\theta}{2}\)

image

Given : \(\lambda=\)1 nC, R = 3m, \(\theta = \pi\) 

\(\vec E_{net}=\dfrac {1×10^{-9}×9×10^9×2}{3}×sin \dfrac{\pi}{2}\)

\(\vec E_{net}=\) 6 N/C

image

Calculate the electric field due to a semicircular portion of ring radius R = 3m having linear charge density  \(\lambda=\)1 nC/m at its center.

image
A

20 N/C

.

B

8 N/C

C

5 N/C

D

6 N/C

Option D is Correct

Electric Field due to a Line of Charge (Rod)

  • Consider a line of charge of length L.
  • To calculate electric field at point P, consider an element at a distance 'x' from point O.
  • Charge on small element dQ = \(\lambda\)dx
  • where, \(\lambda\)= linear charge density,

    dx = length of small element

  • Electric field at point P due to this element 
  •  \(d\vec E=\dfrac {K\lambda\;dx}{r^2}\)

  • From figure 
  • \(tan\;\theta=\dfrac {x}{R}\)

    \(x=R\;tan\;\theta\)

    Differentiating x with respect to \(\theta\)

    \(dx=R\;sec^2\;\theta\;d\theta\)...(i)

  • Also from figure
  •  \(cos\;\theta=\dfrac {R}{r}\)

    \(r=R\;sec\;\theta\)

    \(r^2=R^2\;sec^2\;\theta\) ...(ii)

  • Component of electric field in perpendicular direction to line charge at P,

\(dE_{\perp}=dE\;cos\,\theta\)

\(=\dfrac {K\;\lambda\;dx}{r^2}cos\,\theta\)

  • Component of electric field in parallel direction to line charge at P,

\(dE_{\parallel}=dE\;sin\,\theta\)

\(=\dfrac {K\;\lambda\;dx}{r^2}sin\,\theta\)

  • Total electric field in perpendicular direction due to whole line charge,

\(E_{\perp}=\int dE_{\perp}=\displaystyle\int\dfrac {K\;\lambda\;dx}{r^2}\;cos\;\theta\)

From equation (i) and (ii),

\(E_{\perp}=K\;\lambda =\int\limits_{-\alpha}^{\beta}\dfrac {R\;sec^2\;\theta\;d\theta}{R^2\;sec^2\;\theta}\;cos\;\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\theta\Big]_{-\alpha}^{\beta}\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\beta+sin\;\alpha\Big]\)

  • Total electric field in parallel direction to line charge,

\(E_{\parallel}=\int dE_{\parallel}=\displaystyle\int \dfrac{K\lambda\,dx}{r^2}sin\,\theta\,d\theta\)

\(E_{\parallel}=\int\limits_{-\alpha}^{\beta}\;\dfrac {K\,\lambda\,sin\,\theta(R\,sec^2\,\theta)}{R^2\,sec^2\,\theta} \,d\theta\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\theta \Big]_{-\alpha}^{\beta}\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\beta+cos\;\alpha \Big]\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ cos\;\alpha-cos\,\beta \Big]\)

  • Net electric field at P,

\(\vec E_{net}=\sqrt {(E_{\perp})^2+(E_{\parallel})^2}\)

  • \(\vec E_{net}\) makes an angle \(\theta\) with y-axis, where \(\tan\;\theta=\dfrac {Ex}{Ey}\)

 

  • Now consider a section of ring making an angle \(\alpha\) and \(\beta\) at point P as shown in figure.
  • The section of ring is having a linear charge density (charge per unit length), \(\lambda\).
  • To calculate electric field, consider a small element of length \(d\ell\), making an angle \(d\theta\) and at  \(\ell\)distance from point Q.
  • From Figure

\(\ell=R\,\theta\)

\(d\ell\) \(=R\,d\theta\)

 

 

  • Perpendicular component of Electric field,

\(dE_{\perp}=dE\;cos\,\theta\)

We know, \(dE =\dfrac {K\lambda\;d\ell}{R^2}\)

So, \(dE_{\perp} =\dfrac {K\lambda\;d\ell}{R^2}\;\cos\theta\)

\(dE_{\perp} =\dfrac {K\lambda\;(R\;d\theta\;)cos\theta}{R^2}\;\)      \([\;d\ell=R\;d\theta\;]\)

\(dE_{\perp} =\dfrac {K\lambda\;}{R}\;\cos\theta\;d\theta\)

Integrating both sides,

\(E_{\perp}=\int dE_{\perp}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\cos\;\theta\;d\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R}\, \Big[ sin\;\beta+sin\;\alpha\Big]\)

  • Parallel component of electric field,

\(dE_{\parallel}=dE\;sin\theta\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,d\ell}{R^2}\;sin\theta\)                \(\left [ dE=\dfrac {K\,\lambda\,d\ell}{R^2} \right]\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,(R\;d\theta)}{R^2}\;sin\theta\)         \(\Big[ d\ell =R\,d\theta\Big]\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,\;sin\theta\;d\theta}{R^2}\)

Integrating both sides,

\(E_{\parallel}=\int dE_{\parallel}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\sin\,\theta\;d\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[ cos\;\alpha-cos\;\beta\Big]\)

  • Net electric field due to section of ring,

\(E_{net}=\sqrt{E_{\perp}^2+E_{\parallel}^2}\)

CONCLUSION: From derivation, it is clear that electric field due to section of ring is same as electric field due to finite rod.

Illustration Questions

Calculate the net electric field due to a rod of linear charge density \(\lambda= 5\mu\,\text{C/m}\) on a point lying on a perpendicular bisector at distance R = 3 m and making an angle \(\theta=\) 37° between perpendicular bisector and on end of rod.

A 6 × 103 N/C

B 18 × 103 N/C

C 5 × 103 N/C

D 4 × 107 N/C

×

Component of electric field in perpendicular direction 

\(E_y=\dfrac {K\;\lambda}{R}\big[sin\,\beta+sin\,\alpha\big]\)

where, \(\alpha=\beta=\theta=\)37°

\(E_y=\dfrac {9×10^9×2}{(3)}\;sin\,37°×5×10^{-6}\)

\(E_y=\) 18 × 103 N/C

image

Component of electric field in parallel direction 

\(E_x=\dfrac {K\;\lambda}{R}\big[cos\,\alpha-cos\,\beta\big]\)

where, \(\alpha=\beta=\theta=\)37°

\(E_x=0\)

image

Hence, net electric field 

\(E_{net}=\)18 × 103  N/C

image

Calculate the net electric field due to a rod of linear charge density \(\lambda= 5\mu\,\text{C/m}\) on a point lying on a perpendicular bisector at distance R = 3 m and making an angle \(\theta=\) 37° between perpendicular bisector and on end of rod.

image
A

6 × 103 N/C

.

B

18 × 103 N/C

C

5 × 103 N/C

D

4 × 107 N/C

Option B is Correct

Calculation of Electric Field at a Point due to Combination of Section of Ring and Rod

  • Consider a section of ring of radius R as shown in figure.

  • To calculate electric field at the center of ring due to section of ring, choose the element on the ring.
  • An element of length \(d\ell\) subtending an angle \(d\alpha\) at the center of the ring, as shown in figure.
  • The horizontal components of electric field will cancel out and vertical component will get added, as shown in figure.
  • The net electric field at point O

\(dE_{net}=2dE\, cos\,\alpha\)

Integrating both sides,

\(\int dE_{net}=\int\limits_0^{\theta/2}\,2dE\;cos\alpha\)

\(E_{net}=2\int\limits_0^{\theta/2}\,\dfrac {\lambda\;d\ell}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

  • From figure,

    \(d\ell=R\,d\alpha\)

    \(\vec E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;R\,d\alpha}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)

    \(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R}\Big[sin\;\alpha\Big]_0^{\theta/2}\)

    \(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R} \Big[sin\;\dfrac {\theta}{2}-sin\;0\Big]\)

    \(\vec E_{net}=\dfrac {\lambda}{2\,\pi\,\epsilon_0R}sin\;\dfrac {\theta}{2}\)

     

Electric Field due to a Line of Charge (Rod)

  • Consider a line of charge of length L.
  • To calculate electric field at point P, consider an element at a distance 'x' from point O.
  • Charge on small element dQ = \(\lambda\)dx

where,

 \(\lambda\)= linear charge density

dx = length of small element

  • Electric field at point P due to this element 

 \(d\vec E=\dfrac {K\lambda\;dx}{r^2}\)

  • From figure 

\(tan\;\theta=\dfrac {x}{R}\)

\(x=R\;tan\;\theta\)

Differentiating x with respect to \(\theta\)

\(dx=R\;sec^2\;\theta\;d\theta\)...(i)

  • Also from figure

 \(cos\;\theta=\dfrac {R}{r}\)

\(r=R\;sec\;\theta\)

\(r^2=R^2\;sec^2\;\theta\) ...(ii)

 

  • Component of electric field in perpendicular direction to line charge at P,

\(dE_{\perp}=dE\;cos\,\theta\)

\(=\dfrac {K\;\lambda\;dx}{r^2}cos\,\theta\)

  • Component of electric field in parallel direction to line charge at P,

\(dE_{\parallel}=dE\;sin\,\theta\)

\(=\dfrac {K\;\lambda\;dx}{r^2}sin\,\theta\)

  • Total electric field in perpendicular direction due to whole line charge,

\(E_{\perp}=\int dE_{\perp}=\displaystyle\int\dfrac {K\;\lambda\;dx}{r^2}\;cos\;\theta\)

From equation (i) and (ii),

\(E_{\perp}=K\;\lambda =\int\limits_{-\alpha}^{\beta}\dfrac {R\;sec^2\;\theta\;d\theta}{R^2\;sec^2\;\theta}\;cos\;\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\theta\Big]_{-\alpha}^{\beta}\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\beta+sin\;\alpha\Big]\)

  • Total electric field in parallel direction to line charge,

\(E_{\parallel}=\int dE_{\parallel}=\displaystyle\int \dfrac{K\lambda\,dx}{r^2}sin\,\theta\,d\theta\)

\(E_{\parallel}=\int\limits_{-\alpha}^{\beta}\;\dfrac {K\,\lambda\,sin\,\theta(R\,sec^2\,\theta)}{R^2\,sec^2\,\theta} \,d\theta\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\theta \Big]_{-\alpha}^{\beta}\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\beta+cos\;\alpha \Big]\)

\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ cos\;\alpha-cos\,\beta \Big]\)

  • Net electric field at P,

\(\vec E_{net}=\sqrt {(E_{\perp})^2+(E_{\parallel})^2}\)

  • \(\vec E_{net}\) makes an angle \(\theta\) with y-axis, where \(\tan\;\theta=\dfrac {Ex}{Ey}\)

 

  • Now consider a section of ring making an angle \(\alpha\) and \(\beta\) at point P as shown in figure.
  • The section of ring is having a linear charge density (charge per unit length), \(\lambda\).
  • To calculate electric field, consider a small element of length \(d\ell\), making an angle \(d\theta\) and at \(\ell\) distance from point Q.
  • From Figure,

\(\ell=R\,\theta\)

\(d\ell\) \(=R\,d\theta\)

 

  • Perpendicular component of Electric field,

\(dE_{\perp}=dE\;cos\,\theta\)

We know, \(dE =\dfrac {K\lambda\;d\ell}{R^2}\)

So, \(dE_{\perp} =\dfrac {K\lambda\;d\ell}{R^2}\;\cos\theta\)

\(dE_{\perp} =\dfrac {K\lambda\;(R\;d\theta\;)cos\theta}{R^2}\;\)      \([\;d\ell=R\;d\theta\;]\)

\(dE_{\perp} =\dfrac {K\lambda\;}{R}\;\cos\theta\;d\theta\)

Integrating both sides,

\(E_{\perp}=\int dE_{\perp}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\cos\;\theta\;d\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R}\, \Big[ sin\;\beta+sin\;\alpha\Big]\)

  • Parallel component of electric field,

\(dE_{\parallel}=dE\;sin\,\theta\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,d\ell}{R^2}\;sin\,\theta\)                \(\left [ dE=\dfrac {K\,\lambda\,d\ell}{R^2} \right]\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,(R\;d\theta)}{R^2}\;sin\,\theta\)         \(\Big[ d\ell =R\,d\theta\Big]\)

\(dE_{\parallel}=\dfrac {K\,\lambda\,\;sin\theta\;d\theta}{R^2}\)

Integrating both sides,

\(E_{\parallel}=\int dE_{\parallel}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\sin\,\theta\;d\theta\)

\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[ cos\;\alpha-cos\;\beta\Big]\)

  • Net electric field due to section of ring,

\(E_{net}=\sqrt{E_{\perp}^2+E_{\parallel}^2}\)

CONCLUSION: From derivation, it is clear that electric field due to section of ring is same as electric field due to finite rod.

Illustration Questions

Calculate electric field at point P due to combination of section of ring of radius R = 9 m and semi infinite rod having linear charge density \(\lambda=\) 65 nC/m as shown in figure.

A 65 N/C

B 130 N/C

C 120 N/C

D 45 N/C

×

The given figure is a combination of a section of ring of radius R and a semi-infinite rod having linear charge density. So, net electric field will be as shown in figure.

image

From figure, it is clear that

For rod,  \(\alpha=90°,\;B = 0° , \:\lambda = 65 \; nC/m, \;R = 9\;m\)

For ring,  \(\theta = 90 ° , \; R = 9\; m,\; \lambda = 65 \, nC/m\)

Electric field due to ring  \(\vec{E} _{ring } = \dfrac{2\; k\;\lambda}{R}\;sin \dfrac{\theta }{2}\)

\(\dfrac{2× 9 × 10^9 × 65 × 10 ^{-9} }{9}\; sin\;\dfrac{90^\circ}{2}\)

\(|\vec{E}_{ring}| = 130 \; sin\, 45° \; \\\Rightarrow 65 \sqrt 2\; N/C\)

 

image image

Perpendicular component of electric field due to rod 

\(\vec{E}_{rod\; \bot} = \dfrac{k\; \lambda}{R} \{sin\alpha + sin \beta \}\)

\(\vec{E}_{rod \; \bot} = \dfrac{9 × 10^9 × 65 × 10^{-9}}{9}\; \{sin\;90° + sin \; 0° \}\)

\(|\vec{E}_{rod\; \bot}| = 6 5\; N/C\)

Parallel component of electric field due to rod

\(\vec{E}_{rod_{11}} = \dfrac{k\;\lambda}{R} [cos\;\alpha - cos\,\beta ]\)

\(\vec{E}_{rod_{11}} = \dfrac{9 × 10^9 × 65 × 10^{-9}}{9 }\; \{cos \; 90° - cos\; 0° \}\)

\(\vec{E}_{rod_{11}} = - 65\; N/C\)

\(|\vec{E}_{rod_{11}}| = 65 \; N/C\)

image image

Net electric field in \(x \)-direction 

\(\vec{E}_{netx} = \vec{E}_{ring}\; cos \;45° + E_{rod}\bot\)

\( = 65 \sqrt 2 \; cos \; 45° + 65\)

\(= 65 \sqrt 2 × \dfrac{1}{\sqrt 2} + 65\)

\( =65 +65\\= 130 \; N/C\)

image image

Net electric field in \(y\)– direction 

\(\vec{E}_{nety} = \vec{E}_{ring} \; sin\; 45^\circ - E_{rod_{11}}\)

\( = 65 \sqrt 2 \; sin \; 45^\circ - 65\)

\( = 65 \sqrt 2 × \dfrac{1}{\sqrt 2} - 65\)

 \( = 0\)

 

image image

Thus, resultant electric field due to combination of section of ring and semi-infinite rod

\(\vec{E} _{net} = \sqrt {\vec{E} _{net\,x}+ \vec{E} _{net\,y}}\)

\(\vec{E} _{net} = \sqrt {(130)^2 + (0)^2}\)

\(\vec{E} _{net} = 130 \; N/C\)

image image

Calculate electric field at point P due to combination of section of ring of radius R = 9 m and semi infinite rod having linear charge density \(\lambda=\) 65 nC/m as shown in figure.

image
A

65 N/C

.

B

130 N/C

C

120 N/C

D

45 N/C

Option B is Correct

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