The electric field in the region b-r-a is E-Y4 -0Qr2- Find Y -

Consider a Gaussian spherical surface of radius r #160;where a lt;r lt;b So, the charge inside the surface is Q_in=3Q By Gauss's law, E.4π ...

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Standard XII

Physics

Electric Field Due to a Sphere and Thin Shell

The electric ...

Question

The electric field in the region $b > r > a$ is $E = \frac{Y}{4 π ε_{0}} \frac{Q}{r^{2}}$ . Find $Y$ ?

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Solution

Consider a Gaussian spherical surface of radius r where $a < r < b$
So, the charge inside the surface is $Q_{i n} = 3 Q$
By Gauss's law, $E .4 π r^{2} = \frac{Q_{i n}}{ϵ_{0}} = \frac{3 Q}{ϵ_{0}}$
or $E = \frac{3 Q}{4 π ϵ_{0} r^{2}}$

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The electric field in the region b>r>a is E=Y4πε0Qr2. Find Y ?

Consider a Gaussian spherical surface of radius r where a<r<bSo, the charge inside the surface is Qin=3QBy Gauss's law, E.4πr2=Qinϵ0=3Qϵ0or E=3Q4πϵ0r2

The electric field in the region $b > r > a$ is $E = \frac{Y}{4 π ε_{0}} \frac{Q}{r^{2}}$ . Find $Y$ ?

Consider a Gaussian spherical surface of radius r where $a < r < b$
So, the charge inside the surface is $Q_{i n} = 3 Q$
By Gauss's law, $E .4 π r^{2} = \frac{Q_{i n}}{ϵ_{0}} = \frac{3 Q}{ϵ_{0}}$
or $E = \frac{3 Q}{4 π ϵ_{0} r^{2}}$