Question

A sphere of radius \( R \) has a uniform charge density \( \rho \). A sphere of smaller radius \( \frac{R}{2} \) is cut out from the original sphere, as shown in the figure. The center of the cut-out sphere lies at \( z = \frac{R}{2} \). After the smaller sphere has been cut out, the magnitude of the electric field at \( z = -\frac{R}{2} \) is \( \frac{\rho R}{n \epsilon_0} \). The value of the integer \( n \) is: 

Hint:

For problems involving charge distributions and symmetry, using Gauss's law can greatly simplify the process, especially when dealing with spherical symmetry.

Answer 1

Step 1: Understanding the problem.
We have a charged sphere with charge density \( \rho \), and a smaller sphere is cut out from it. The question asks for the electric field at a point due to the charge distribution, considering the symmetry of the setup.
Step 2: Applying Gauss's Law.
We can use Gauss’s law to solve for the electric field at a point outside the charged sphere. First, we know that the electric field due to a uniformly charged sphere is given by: \[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q_{\text{enc}}}{r^2} \] where \( Q_{\text{enc}} \) is the enclosed charge, and \( r \) is the distance from the center. After the smaller sphere is cut out, we can treat the remaining charge as a net charge distribution.
Step 3: Considering the cut-out sphere.
The electric field at \( z = -\frac{R}{2} \) is due to the charge left in the sphere after cutting out the smaller sphere. The magnitude of the electric field is related to the net charge left after the cut-out.
Step 4: Conclusion.
By solving the equations, the value of \( n \) is determined to be 6. Thus, the electric field at \( z = -\frac{R}{2} \) is \( \frac{\rho R}{6 \epsilon_0} \).