Chapter 22, Problem 83P

(a)

To determine

To Show:The electric field inside a solid sphere having a charge density $\rho$ at a distance r is $\frac{\rho r}{3{\epsilon }_0}\widehat{r}$

(b)

To determine

The electric field at the given points 1 and 2 as shown in the figure.

Given:

The radius of the cavity is $\frac{R}{2}$ .

Explanation:

Draw a diagram to show the position of the cavity.

  

The electric field inside a cavity is equal to the sum of the electric field due to the original uncut sphere and the electric field due to the sphere of the cavity but with a uniform negative charge density.

Consider a point P at position 1.

Write the expression for the net electric field at position 1.

  $E_1=E_P^R+E_P^b$

Here, $E_1$ is the electric field at position 1, $E_P^R$ is the electric field at $\text{P}$ due to the uncut sphere of radius $R$ and $E_P^b$ is the electric field at P due to the cavity.

Substitute $\frac{\rho R}{3{\epsilon }_0}$ for $E_P^R$ and $\frac{-\rho b}{3{\epsilon }_0}$ for $E_P^b$ in the above expression.

  $\begin{array}{c}{E}_1=\frac{\rho \text{}}{3{\epsilon }_0}(R-b)\\ =\frac{\rho \text{}}{3{\epsilon }_0}(R-\frac{R}{2})\\ =\frac{\rho \text{}R}{6{\epsilon }_0}\end{array}$

The direction will be radially outward.

Take the centrepoint $\text{O}$ at position 2.The electric field will be only due to the cavity as the electric field due to the uncut sphere at the centre will be zero.

Write the expression for the electric field at position 2.

  $E_2=E_P^R+E_P^b$

Here, $E_2$ is the electric field at position 2.

Substitute 0 for $E_P^R$ and $\frac{-\rho b}{3{\epsilon }_0}$ for $E_P^b$ in the above expression.

  $\begin{array}{c}{E}_2=\frac{-\rho b}{3{\epsilon }_0}\\ =-\frac{\rho R}{6{\epsilon }_0}\end{array}$

Negative sign indicates that the electric field will be towards the centre of the cavity.

Conclusion:

Thus, the electric field at point 1 is $\frac{\rho \text{}R}{6{\epsilon }_0}$ and the electric field at point 2 is $-\frac{\rho R}{6{\epsilon }_0}$

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