1. The problem asks to describe the concept of electric dipole with an illustration. An electric dipole consists of two equal and opposite charges, +q and -q, separated by a small distance \( \mathbf{d} \). The dipole moment \( \mathbf{p} \) is a vector defined as: $$\mathbf{p} = q \mathbf{d}$$ where \( \mathbf{d} \) points from the negative to the positive charge. An illustration typically shows two opposite charges with an arrow from -q to +q marking \( \mathbf{p} \). 2. Next, we explain an equipotential surface with an example. An equipotential surface is a surface on which the electric potential \( V \) is constant everywhere. For example, around a point charge, equipotential surfaces are concentric spheres centered on the charge. 3. The concept of spherical symmetry with a single point charge means that the electric field and potential depend only on the distance \( r \) from the charge, not on the direction. Mathematically, \( V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \) depends only on \( r \). This symmetry implies fields are spherically symmetric. 4. For multiple point charges, spherical symmetry generally does not hold because potential and field depend on position vectors to each charge independently. Hence, no spherical symmetry unless the charge configuration is symmetrical in all directions. 5. We draw the Electric Flux Density \( D \) versus Radius \( R \) graph for a spherical shell charge using Gauss's law. Given: a = 1 m, b = 2 m, surface charge density \( \sigma = 1 \ C/m^2 \). The piecewise definition is: - For \( R < a \): \( D = 0 \) (inside shell) - For \( a \leq R \leq b \): \( D = \frac{1}{R^2} \) - For \( R > b \): \( D = 0 \) (outside shell) This is because the charge is distributed on the shell; inside and outside the shell have zero \( D \) field. 6. Explanation of the graph: The electric flux density is zero inside the inner radius since no enclosed charge there. Between \( a \) and \( b \), \( D \) follows \( \frac{1}{R^2} \) as the flux emanates from the shell surface. Outside \( b \), \( D \) is zero because the shell charge acts like a shell—no field outside. 7. Two dipoles with moments:\ \( \mathbf{p_1} = -5 \mathbf{a} \) nC·m at (0,0,-2)\ \( \mathbf{p_2} = 9 \mathbf{a} \) nC·m at (0,0,3)\ Find the potential at origin. The potential \( V \) due to a dipole at position \( \mathbf{r'} \) is: $$V = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$$ where \( \mathbf{r} = \mathbf{r_{obs}} - \mathbf{r'} \) is vector from dipole to origin. Calculate: \( \mathbf{r_1} = \mathbf{0} - (0,0,-2) = (0,0,2) \), \( r_1 = 2 \) \( \mathbf{r_2} = \mathbf{0} - (0,0,3) = (0,0,-3) \), \( r_2 = 3 \) Assuming \( \mathbf{a} = \hat{z} \) direction, $$ V_1 = \frac{1}{4 \pi \epsilon_0} \frac{-5 \hat{z} \cdot (0,0,2)}{2^3} = \frac{1}{4 \pi \epsilon_0} \frac{-10}{8} = \frac{-5/4}{4 \pi \epsilon_0} $$ $$ V_2 = \frac{1}{4 \pi \epsilon_0} \frac{9 \hat{z} \cdot (0,0,-3)}{3^3} = \frac{1}{4 \pi \epsilon_0} \frac{-27}{27} = \frac{-1}{4 \pi \epsilon_0} $$ Total potential: $$ V = V_1 + V_2 = \frac{1}{4 \pi \epsilon_0} \left(-\frac{5}{4} - 1\right) = \frac{-9/4}{4 \pi \epsilon_0} $$ 8. Point charges given: \( q_1=1 \) nC at (0,0,0) \( q_2=4 \) nC at (0,0,1) \( q_3=3 \) nC at (1,0,0) 9. The energy of the system is the sum of potential energies between pairs: $$ U = \frac{1}{4 \pi \epsilon_0} \sum_{i