Concentric conducting spheres are located at \(r=5 \mathrm{~mm}\) and \(r=20 \mathrm{~mm}\) The region between the spheres is filled with a perfect dielectric. If the inner sphere is at \(100 \mathrm{~V}\) and the outer sphere is at \(0 \mathrm{~V}(a)\) Find the location of the \(20 \mathrm{~V}\) equipotential surface. \((b)\) Find \(E_{r, \max } \cdot(c)\) Find \(\epsilon_{r}\) if the surface charge density on the inner sphere is \(1.0 \mu \mathrm{C} / \mathrm{m}^{2}\).

First, let's consider the electric field in the region between the spheres. Since concentric conducting spheres are equipotential surfaces, we can use the formula for the electric field in a radial direction: \(E_r = \frac{kQ}{r^2}\) To find the constant k, we can use the information that the potential of the inner sphere is 100 V and the potential of the outer sphere is 0 V. We know for concentric spheres: \(V(r) = V_o - \frac{kQ}{r}\), where r is the radius. We can set up the following system of equations: $\begin{cases} 0 = V_o - \frac{kQ}{20} \\ 100 = V_o - \frac{kQ}{5} \end{cases}$ Solve for k in terms of V_o: \(kQ = 5V_o\) Now plug this back into the first equation: \(0 = V_o - \frac{5V_o}{20}\) Solving for Vo and kQ gives: \(V_o = 20 V\) and \(kQ = 100 V\) Now we have the electric field in the radial direction: \(E_r = \frac{100V}{r^2}\)

Next, we'll find the location of the equipotential surface with a potential of 20 V. We can use the potential formula mentioned before: \(20V = V_o - \frac{100V}{r}\) Plugging in the value for V_o, we have: \(20V = 20V - \frac{100V}{r}\) Solving for r: \(r = 5mm\) So, the location of the 20 V equipotential surface is at \(r = 5mm\).

To find the maximum of the radial electric field, we consider the fact that the electric field is strongest near the inner sphere. Thus, the maximum electric field will occur at r = 5mm: \(E_{r,max} = \frac{100V}{(5mm)^2}\) Converting to meters: \(E_{r,max} = \frac{100V}{(0.005m)^2}\) \(E_{r,max} = 4 × 10^6 V/m\)

We are given the surface charge density (σ) of the inner sphere as \(1.0 μC/m^2\). We can relate this to the electric field through the following equation, where ε is the absolute permittivity, and εr is the relative permittivity: \(σ = εE_r\) So after converting into standard units and rearranging for εr, we get: \(ε_r = \frac{σ}{ε_0 E_{r,max}}\) \(ε_r = \frac{1.0 × 10^{-6} C/m^2}{(8.85 × 10^{-12} C^2/Nm^2)(4 × 10^6 V/m)}\) \(ε_r ≈ 2.8\) Thus, the relative permittivity of the dielectric (εr) is approximately 2.8.