A long round dielectric cylinder is polarized so that the vector P - alpha r- where alpha is a positive constant and r is the distance from the axis- Find the space density rho- of bound charges as a of distance r from the axis-

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Given -vec -P- - -alpha -vec -r- where -vec -r- - distance from the axis- The space density of charges is given by- -rho- - -div -vec -P- - -2-alpha-On using- -div -vec -r- - -dfrac -1-r- -dfrac -partial-partial r- -vec -r- - -vec -r- - 2-

A long round dielectric cylinder is polarized so that the vector $$P = \alpha r$$, where $$\alpha$$ is a positive constant and $$r$$ is the distance from the axis. Find the space density $$\rho'$$ of bound charges as a function of distance $$r$$ from the axis.

Given $$\vec {P} = \alpha \vec {r}$$, where $$\vec {r} =$$ distance from the axis. The space density of charges is given by, $$\rho' = -div \vec {P} = -2\alpha$$
On using. $$\div \vec {r} = \dfrac {1}{r} \dfrac {\partial}{\partial r} (\vec {r} . \vec {r}) = 2$$.

A long round dielectric cylinder is polarized so that the vector $$P = \alpha r$$, where $$\alpha$$ is a positive constant and $$r$$ is the distance from the axis. Find the space density $$\rho'$$ of bound charges as a function of distance $$r$$ from the axis.

Given $$\vec {P} = \alpha \vec {r}$$, where $$\vec {r} =$$ distance from the axis. The space density of charges is given by, $$\rho' = -div \vec {P} = -2\alpha$$On using. $$\div \vec {r} = \dfrac {1}{r} \dfrac {\partial}{\partial r} (\vec {r} . \vec {r}) = 2$$.

Given $$\vec {P} = \alpha \vec {r}$$, where $$\vec {r} =$$ distance from the axis. The space density of charges is given by, $$\rho' = -div \vec {P} = -2\alpha$$
On using. $$\div \vec {r} = \dfrac {1}{r} \dfrac {\partial}{\partial r} (\vec {r} . \vec {r}) = 2$$.