Question

A solid spherical shere of surface charge density o, charge q and radius R. Find the Electric potential Inside and Outside the shere respectively?​

Answer 1

Answer:

Consider a charged sphere with a symmetrical distribution of charge. Gauss' Law tells us that the electric field outside the sphere is the same as that from a point charge. This implies that outside the sphere the potential also looks like the potential from a point charge.

Explanation:

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Answer 2

Electric potential :

Explanation:

• Let a thin spherical shell of radius R with a positive charge q be distributed uniformly on the surface.
• As the charge is constantly distributed, the electric field is symmetrical and directed radially outward.

1)  Electric potential inside the sphere:

• For the gaussian surface concentric with the shell of radius r, (r>R)

        $\oint$ E ds = E ( 4π$r^2$)

• According to the gauss law,

        E ( 4π$r^2$) = $\displaystyle \frac{Q}{\epsilon_0}$

• Since the charge enclosed inside the spherical shell is zero.

         E = 0

Hence, the electric potential due to a uniform charge inside the spherical shell is zero at all points.

2)   Electric potential outside the sphere:

• Consider (r>R).

• Draw a spherical surface of radius r.

• According to the gauss law,

       $\oint$ E ds =  $\displaystyle \frac{Q}{\ q_0}$

• While $\bar_E$ is perpendicular to the gaussian surface, the angle between $\bar_E$  is 0.  $\bar_E$ being constant can be taken out of integral.

        E ( 4π$r^2$) = $\displaystyle \frac{q}{\epsilon_0}$

Thus electric field outside a uniform charge spherical shell is the same as if all the charge q were concentrated as a point charge at the center of the shell.