Learn calculation of electric field due to uniformly charged ring and disk along the central axis, practice electric field continuous charge distribution.
Case 1 : Thin Rod
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
Charge on the small element \( dQ = dx ×\text{ charge density}\)
\(dQ=dx×\dfrac {Q}{2\pi R}\)
Case 3 : 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).
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
\(\vec {dE}=\dfrac {K\cdot dQ}{r^2}\)
\(\vec {dE}=\dfrac {K\cdot dQ}{\sqrt {R^2+a^2}}\)
So, Electric field in x-direction
\(dE_x=\dfrac {K\cdot dQ}{R^2+a^2}\,cos\,\theta\)
\(\vec E=\int \dfrac {K\cdot dQ}{R^2+a^2}cos\;\theta\; \hat i\)
For cos \(\theta\) :
\(cos\,\theta=\dfrac {a}{\sqrt {R^2+a^2}}\)
\(\vec E=\int dE_x\,\hat i=\int \dfrac {K.dQ}{(R^2+a^2)}\cdot\dfrac {a}{\sqrt {R^2+a^2}}\,\hat i\)
\(\vec E=E_x\,\hat i=\dfrac {Ka\int dQ}{(R^2+a^2)^{3/2}}\,\hat i\)
\(\vec E=\dfrac {K\,Q\,a}{(R^2+a^2)^{3/2}}\,\hat i\)
\(\vec E=\dfrac {K\,Q\,a}{(R^2+a^2)^{3/2}}\,\hat r\)
where, \(\hat r\) is the unit vector along the axis of ring.
where, \(\lambda\)= linear charge density,
dx = length of small element
\(d\vec E=\dfrac {K\lambda\;dx}{r^2}\)
\(x=R\;tan\;\theta\)
Differentiating x with respect to \(\theta\)
\(dx=R\;sec^2\;\theta\;d\theta\)...(i)
\(cos\;\theta=\dfrac {R}{r}\)
\(r=R\;sec\;\theta\)
\(r^2=R^2\;sec^2\;\theta\) ...(ii)
\(dE_{\perp}=dE\;cos\,\theta\)
\(=\dfrac {K\;\lambda\;dx}{r^2}cos\,\theta\)
\(dE_{\parallel}=dE\;sin\,\theta\)
\(=\dfrac {K\;\lambda\;dx}{r^2}sin\,\theta\)
\(E_{\perp}=\int dE_{\perp}=\displaystyle\int\dfrac {K\;\lambda\;dx}{r^2}\;cos\;\theta\)
From equation (i) and (ii),
\(E_{\perp}=K\;\lambda =\int\limits_{-\alpha}^{\beta}\dfrac {R\;sec^2\;\theta\;d\theta}{R^2\;sec^2\;\theta}\;cos\;\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\theta\Big]_{-\alpha}^{\beta}\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\beta+sin\;\alpha\Big]\)
\(E_{\parallel}=\int dE_{\parallel}=\displaystyle\int \dfrac{K\lambda\,dx}{r^2}sin\,\theta\,d\theta\)
\(E_{\parallel}=\int\limits_{-\alpha}^{\beta}\;\dfrac {K\,\lambda\,sin\,\theta(R\,sec^2\,\theta)}{R^2\,sec^2\,\theta} \,d\theta\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\theta \Big]_{-\alpha}^{\beta}\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\beta+cos\;\alpha \Big]\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ cos\;\alpha-cos\,\beta \Big]\)
\(\vec E_{net}=\sqrt {(E_{\perp})^2+(E_{\parallel})^2}\)
\(\ell=R\,\theta\)
\(d\ell\) \(=R\,d\theta\)
\(dE_{\perp}=dE\;cos\,\theta\)
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,
\(E_{\perp}=\int dE_{\perp}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\cos\;\theta\;d\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R}\, \Big[ sin\;\beta+sin\;\alpha\Big]\)
\(dE_{\parallel}=dE\;sin\theta\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,d\ell}{R^2}\;sin\theta\) \(\left [ dE=\dfrac {K\,\lambda\,d\ell}{R^2} \right]\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,(R\;d\theta)}{R^2}\;sin\theta\) \(\Big[ d\ell =R\,d\theta\Big]\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,\;sin\theta\;d\theta}{R^2}\)
Integrating both sides,
\(E_{\parallel}=\int dE_{\parallel}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\sin\,\theta\;d\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[ cos\;\alpha-cos\;\beta\Big]\)
\(E_{net}=\sqrt{E_{\perp}^2+E_{\parallel}^2}\)
CONCLUSION: From derivation, it is clear that electric field due to section of ring is same as electric field due to finite rod.
A 6 × 103 N/C
B 18 × 103 N/C
C 5 × 103 N/C
D 4 × 107 N/C
\(dE_{net}=2dE\, cos\,\alpha\)
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,
\(d\ell=R\,d\alpha\)
\(\vec E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;R\,d\alpha}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)
\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R}\Big[sin\;\alpha\Big]_0^{\theta/2}\)
\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R} \Big[sin\;\dfrac {\theta}{2}-sin\;0\Big]\)
\(\vec E_{net}=\dfrac {\lambda}{2\,\pi\,\epsilon_0R}sin\;\dfrac {\theta}{2}\)
where,
\(\lambda\)= linear charge density
dx = length of small element
\(d\vec E=\dfrac {K\lambda\;dx}{r^2}\)
\(tan\;\theta=\dfrac {x}{R}\)
\(x=R\;tan\;\theta\)
Differentiating x with respect to \(\theta\)
\(dx=R\;sec^2\;\theta\;d\theta\)...(i)
\(cos\;\theta=\dfrac {R}{r}\)
\(r=R\;sec\;\theta\)
\(r^2=R^2\;sec^2\;\theta\) ...(ii)
\(dE_{\perp}=dE\;cos\,\theta\)
\(=\dfrac {K\;\lambda\;dx}{r^2}cos\,\theta\)
\(dE_{\parallel}=dE\;sin\,\theta\)
\(=\dfrac {K\;\lambda\;dx}{r^2}sin\,\theta\)
\(E_{\perp}=\int dE_{\perp}=\displaystyle\int\dfrac {K\;\lambda\;dx}{r^2}\;cos\;\theta\)
From equation (i) and (ii),
\(E_{\perp}=K\;\lambda =\int\limits_{-\alpha}^{\beta}\dfrac {R\;sec^2\;\theta\;d\theta}{R^2\;sec^2\;\theta}\;cos\;\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\theta\Big]_{-\alpha}^{\beta}\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[sin\;\beta+sin\;\alpha\Big]\)
\(E_{\parallel}=\int dE_{\parallel}=\displaystyle\int \dfrac{K\lambda\,dx}{r^2}sin\,\theta\,d\theta\)
\(E_{\parallel}=\int\limits_{-\alpha}^{\beta}\;\dfrac {K\,\lambda\,sin\,\theta(R\,sec^2\,\theta)}{R^2\,sec^2\,\theta} \,d\theta\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\theta \Big]_{-\alpha}^{\beta}\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ -cos\,\beta+cos\;\alpha \Big]\)
\(E_{\parallel}=\dfrac {K\,\lambda}{R}\Big[ cos\;\alpha-cos\,\beta \Big]\)
\(\vec E_{net}=\sqrt {(E_{\perp})^2+(E_{\parallel})^2}\)
\(\ell=R\,\theta\)
\(d\ell\) \(=R\,d\theta\)
\(dE_{\perp}=dE\;cos\,\theta\)
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,
\(E_{\perp}=\int dE_{\perp}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\cos\;\theta\;d\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R}\, \Big[ sin\;\beta+sin\;\alpha\Big]\)
\(dE_{\parallel}=dE\;sin\,\theta\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,d\ell}{R^2}\;sin\,\theta\) \(\left [ dE=\dfrac {K\,\lambda\,d\ell}{R^2} \right]\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,(R\;d\theta)}{R^2}\;sin\,\theta\) \(\Big[ d\ell =R\,d\theta\Big]\)
\(dE_{\parallel}=\dfrac {K\,\lambda\,\;sin\theta\;d\theta}{R^2}\)
Integrating both sides,
\(E_{\parallel}=\int dE_{\parallel}=\dfrac {K\;\lambda}{R} \int\limits_{-\alpha}^{\beta}\sin\,\theta\;d\theta\)
\(E_{\perp}=\dfrac {K\;\lambda}{R} \Big[ cos\;\alpha-cos\;\beta\Big]\)
\(E_{net}=\sqrt{E_{\perp}^2+E_{\parallel}^2}\)
CONCLUSION: From derivation, it is clear that electric field due to section of ring is same as electric field due to finite rod.
\(dq =\sigma\,dA\)
\(dq =\sigma(2\pi r\,dr)\)
\(dq =2\pi\,\sigma\, r\,dr\)
\(d\vec E=\dfrac {K\, dq}{r^2+x^2}\)
\(d\vec E_x=\dfrac {K}{(r^2+x^2)}\,cos\,\theta (2\pi r\,dr\,\sigma)\)
\(d\vec E_x=\dfrac {K}{(r^2+x^2)}×\dfrac {x}{\sqrt {x^2+r^2}}\,(2\pi\,\sigma\, r\,dr\,)\) \(\left [ \because cos\theta=\dfrac {x}{\sqrt {x^2+r^2}} \right]\)
\(d\vec E_x=\dfrac {K\,x\,(2\pi\,\sigma\, r\,dr\,)}{(r^2+x^2)^{3/2}}\)
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, \(\vec E=2\,\pi Kx\sigma \displaystyle\int\limits_0^R\dfrac{dt}{2t^{3/2}}\hat i\)
\(\vec E=\dfrac{2\,\pi Kx\sigma}{2} \displaystyle\int\limits_0^R{t^{-3/2}dt}\,\hat i\)
\(\vec E=\pi\,Kx\sigma \Bigg[\dfrac{t^{-3/2+1}}{{-3/2}+1}\Bigg]_0^R\,\hat i\)
\(\vec E=\pi Kx\sigma \times \Bigg[\dfrac{-2}{\sqrt t}\Bigg]_0^R\,\,\hat i\)
\(\vec E=2\pi K x\sigma \Bigg[\dfrac{-1}{\sqrt{r^2+x^2}}\Bigg]_0^R \,\hat i\)
\(\vec E=2\pi Kx\,\sigma\Bigg[\dfrac{-1}{\sqrt{R^2+x^2}}+\dfrac{1}{x}\Bigg] \,\hat i\\ \vec E=2\pi K\sigma\Bigg[1-\dfrac{x}{\sqrt{R^2+x^2}}\Bigg]\,\hat i\)
\(\vec E=2\pi\, K\,\sigma\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)
\(=\dfrac {2\pi\,\sigma}{4\,\pi\epsilon_0}\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)
\(=\dfrac {\sigma}{2\,\epsilon_0}\;\left [ 1-\dfrac{x}{\sqrt {R^2+x^2}}\right]\hat r\)
where, \(\hat r\)is the unit vector along axis.
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
\(dE_{net}=2d\,E cos\,\alpha\)
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,
\(d\ell=R\,d\alpha\)
\(\vec E_{net}=2\displaystyle\int\limits_0^{\theta/2}\,\dfrac {\lambda\;R\,d\alpha}{4\,\pi\,\epsilon_0R^2}\;cos\alpha\)
\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R}\Big[sin\;\alpha\Big]_0^{\theta/2}\)
\(\vec E_{net}=\dfrac {2\lambda}{4\,\pi\,\epsilon_0R} \Big[sin\;\dfrac {\theta}{2}-sin\;0\Big]\)
\(\vec E_{net}=\dfrac {\lambda}{2\,\pi\,\epsilon_0R}sin\;\dfrac {\theta}{2}\)