Q18P - Introduction to Electrodynamics

Question

Calculate the divergence of the following vector functions:

Two spheres, each of radius R and carrying uniform volume charge densities +p and -p , respectively, are placed so that they partially overlap (Fig. 2.28). Call the vector from the positive center to the negative center d. Show that the field in the region of overlap is constant, and find its value. [Hint: Use the answer to Prob. 2.12.]

Step-by-Step Solution

Answer

The electric field in the overlapped region is $E=\frac{p}{3{\epsilon }_0}d.$

Step 1: Determine the expression for the electric field in the two sphere.

Consider the diagram for the sphere that has radius R and carries the uniform charge such that they are overlapped with the distance from the center as d.

                                 

Consider the equation for electric field inside the positive sphere is,

 ${\mathbf{E}}_{\mathbf{+}}\mathbf{=}\frac{\mathbf{p}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{r}}_{\mathbf{+}}$

 

Here, r is the radius of the positive centered sphere, p is the uniform charge density.

 

Consider the equation for electric field inside the negative sphere is,

${\mathbf{E}}_{-}\mathbf{=}\frac{\mathbf{p}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{r}}_{-}$ 

 

Here, $r_{-}$  is the radius of the positive centered sphere.

Step 2: Solve for the electric filed in the overlapping region.

Consider the expression for the resultant electric field as,

 $E=E_{+}+E_{-}$

 

Substitute  $\frac{p}{3{\epsilon }_0}{r}_{+}$for $E_{+}$and $\frac{p}{3{\epsilon }_0}{r}_{-}$ for $E_{+}$ in the equation.

$E=\frac{p}{3{\epsilon }_0}{r}_{+}-\frac{p}{3{\epsilon }_0}{r}_{-}$ 

 

From the diagram $d=r_{+}-r_{-}$ .

 

Solve further as,

$E=\frac{p}{3{\epsilon }_0}d$ 

 

Therefore, the electric field in the overlapped region is $E=\frac{p}{3{\epsilon }_0}d$ .