A thin disk of radius b= $ 2a $ has a concentric hole of radius a in it. It carries uniform surface charge on it. If the electric field on its axis at height h ( h$<<$a) from its center is given by ‘cg’, then value of c is:
A. $ \dfrac{\sigma }{a\varepsilon \circ } $
B. $ \dfrac{\sigma }{2a\varepsilon \circ } $
C. $ \dfrac{\sigma }{4a\varepsilon \circ } $
D. $ \dfrac{\sigma }{8a\varepsilon \circ } $
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Hint: The formula of electric field due to uniform changed disk is given by E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ \dfrac{1-h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $
We can find electric field at distance ‘h’ by subtracting the electric field due to disk of radius a from disk of radius ‘2a’.On applying the given condition and putting $ E=ch, $ we can calculate h.
Complete step-by-step
Electric field due to uniform changed disk with surface charge density $ \sigma $ is given by:
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1=\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $ where h is the Distance along the axis of disk from the center.
Electric field at distance can be calculated by subtracting the electric field due to disk of radius from disk of radius
So we get
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{\left( 2a \right)}^{2}}}} \right]\text{ }-\dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{a}^{2}}}} \right) $
as h << a, $ \sqrt{{{h}^{2}}+{{\left( 2a \right)}^{2}}}=2a\text{ }and\text{ }\sqrt{{{h}^{2}}+{{a}^{2}}}=a $
Hence, the question becomes
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{2a} \right)\text{ }-\dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{a} \right) $
= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\dfrac{\sigma h}{4\varepsilon \circ }\text{ }-\text{ }\dfrac{\sigma }{2\varepsilon \circ }\text{ +}\dfrac{\sigma h}{2\varepsilon \circ a} $
= $ \dfrac{\sigma h}{4\varepsilon \circ a} $
Now, as given in the question
E= ch
So,
Ch= $ \dfrac{\sigma h}{4\varepsilon \circ a} $
C= $ \dfrac{\sigma }{4\varepsilon \circ } $ .
Note
The electric field of a disc of charge can be found by superposing the point charge elements.
This can be facilitated by summing the field of charged rings. The integral over the charged disc takes the form
E=k $ \sigma 2\pi h\int{\dfrac{{{r}^{1}}d{{r}^{1}}}{\circ {{\left( {{h}^{2}}+{{r}^{12}} \right)}^{3/2}}}}-a $
Here h=perpendicular distance from center of disk to the point an axio.
R=radius of inner concentric axio.
R=radius of outer concentric axio.
On integrating question a, we get
E=k $ \sigma 2\pi \left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $
As K= $ \dfrac{1}{4\varepsilon \circ }\text{ }=coulomb's\text{ constant} $
Hence
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $ .
We can find electric field at distance ‘h’ by subtracting the electric field due to disk of radius a from disk of radius ‘2a’.On applying the given condition and putting $ E=ch, $ we can calculate h.
Complete step-by-step
Electric field due to uniform changed disk with surface charge density $ \sigma $ is given by:
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1=\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $ where h is the Distance along the axis of disk from the center.
Electric field at distance can be calculated by subtracting the electric field due to disk of radius from disk of radius
So we get
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{\left( 2a \right)}^{2}}}} \right]\text{ }-\dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{a}^{2}}}} \right) $
as h << a, $ \sqrt{{{h}^{2}}+{{\left( 2a \right)}^{2}}}=2a\text{ }and\text{ }\sqrt{{{h}^{2}}+{{a}^{2}}}=a $
Hence, the question becomes
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{2a} \right)\text{ }-\dfrac{\sigma }{2\varepsilon \circ }\text{ }\left( 1-\dfrac{h}{a} \right) $
= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\dfrac{\sigma h}{4\varepsilon \circ }\text{ }-\text{ }\dfrac{\sigma }{2\varepsilon \circ }\text{ +}\dfrac{\sigma h}{2\varepsilon \circ a} $
= $ \dfrac{\sigma h}{4\varepsilon \circ a} $
Now, as given in the question
E= ch
So,
Ch= $ \dfrac{\sigma h}{4\varepsilon \circ a} $
C= $ \dfrac{\sigma }{4\varepsilon \circ } $ .
Note
The electric field of a disc of charge can be found by superposing the point charge elements.
This can be facilitated by summing the field of charged rings. The integral over the charged disc takes the form
E=k $ \sigma 2\pi h\int{\dfrac{{{r}^{1}}d{{r}^{1}}}{\circ {{\left( {{h}^{2}}+{{r}^{12}} \right)}^{3/2}}}}-a $
Here h=perpendicular distance from center of disk to the point an axio.
R=radius of inner concentric axio.
R=radius of outer concentric axio.
On integrating question a, we get
E=k $ \sigma 2\pi \left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $
As K= $ \dfrac{1}{4\varepsilon \circ }\text{ }=coulomb's\text{ constant} $
Hence
E= $ \dfrac{\sigma }{2\varepsilon \circ }\text{ }\left[ 1-\dfrac{h}{\sqrt{{{h}^{2}}+{{r}^{2}}}} \right] $ .