Let's solve the problem step by step: --- ## **Step 1: Given Data** - Spherically symmetric charge distribution: \(\rho_v = \frac{1}{r}\) C/m³ within a sphere of radius \(R = 2\) m. - Center of sphere: origin \((0,0,0)\) - Conducting plane at \(z = -10\), held at zero potential. - Point charge \(q = 5\) nC \(= 5 \times 10^{-9}\) C to be moved from infinity to \((0, 0, -5)\). - **Hint:** Use image theory. --- ## **Step 2: Find Potential at (0, 0, -5)** The work required to bring the charge from infinity to \((0,0,-5)\) is: \[ W = qV(0,0,-5) \] where \(V(0,0,-5)\) is the potential at \((0,0,-5)\) due to all sources (the sphere, and the image charge). ### **A. Potential Due to Volume Charge Distribution** First, calculate the total charge inside the sphere: \[ Q_{sphere} = \int_{0}^{R} \rho_v \cdot 4\pi r^2 dr = \int_{0}^{2} \frac{1}{r} \cdot 4\pi r^2 dr = 4\pi \int_{0}^{2} r \, dr = 4\pi \left[ \frac{r^2}{2} \right]_0^2 = 4\pi \times 2 = 8\pi \text{ C} \] Now, find the potential at \((0,0,-5)\) due to this charge distribution. For simplicity, since the point is outside the sphere, we can treat the sphere as a point charge at the origin (by Gauss's law, for spherically symmetric distributions): \[ V_{sphere} = \frac{1}{4\pi \varepsilon_0} \frac{Q_{sphere}}{r} \] where \(r\) is the distance from the origin to \((0,0,-5)\): \[ r = \sqrt{0^2 + 0^2 + (-5)^2} = 5 \] So, \[ V_{sphere} = \frac{1}{4\pi \varepsilon_0} \frac{8\pi}{5} = \frac{2}{\varepsilon_0 \cdot 5} \] --- ### **B. Potential Due to Image Charge** When a point charge \(q\) is placed at \((0, 0, -5)\), an image charge of \(-q\) is placed at \((0, 0, -15)\) (since the plane is at \(z=-10\)). The potential at \((0, 0, -5)\) **due to the image charge** is: \[ V_{image} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{|z_1 - z_2|} \] where \(z_1 = -5\), \(z_2 = -15\): \[ |z_1 - z_2| = |-5 - (-15)| = 10 \] So, \[ V_{image} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{10} \] --- ### **C. Potential Due to the Sphere at (0,0,-5)** Since the sphere's charge is fixed and independent of the new point charge, its contribution is simply the value calculated above. --- ## **Step 3: Total Potential at (0,0,-5)** Sum the contributions: \[ V_{total} = V_{sphere} + V_{image} \] But **the work done** is only due to external sources (the sphere and the conducting plane), not the self-energy of the point charge. Thus, **only include the potential from the sphere and the image charge** at the location due to other sources: \[ V_{other} = V_{sphere} + V_{image} \] But note, \(V_{image}\) is actually the potential at \((0,0,-5)\) due to the image of \(q\) itself, so when bringing \(q\) from infinity, this potential difference is what matters along with the sphere. \[ V_{other} = \frac{1}{4\pi \varepsilon_0} \frac{8\pi}{5} + \frac{1}{4\pi \varepsilon_0} \frac{-q}{10} \] But, the potential due to the image charge is for the *self-energy*, so the work done by external agent should only consider the sphere and the effect of the plane on \(q\). But let's clarify: **the work includes both the sphere and image charge**. So, \[ W = q \left( V_{sphere} + V_{image} \right) \] ## **Step 4: Plug in the Numbers** Let’s plug in the values: - \(q = 5 \times 10^{-9}\) C - \(\varepsilon_0 = 8.854 \times 10^{-12}\) F/m #### **Calculate \(V_{sphere}\):** \[ V_{sphere} = \frac{2}{\varepsilon_0 \cdot 5} \] #### **Calculate \(V_{image}\):** \[ V_{image} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{10} \] --- ### **Numerical Calculation** #### **A. \(V_{sphere}\):** \[ V_{sphere} = \frac{2}{5 \times 8.854 \times 10^{-12}} = \frac{2}{44.27 \times 10^{-12}} = 4.52 \times 10^{10} \text{ V} \] #### **B. \(V_{image}\):** First, calculate \(k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\) N·m²/C². \[ V_{image} = 8.99 \times 10^9 \cdot \frac{-5 \times 10^{-9}}{10} = 8.99 \times 10^9 \cdot -0.5 \times 10^{-9} = -4.495 \text{ V} \] #### **C. Total Work** \[ W = q (V_{sphere} + V_{image}) \] \[ W = 5 \times 10^{-9} \left(4.52 \times 10^{10} - 4.495\right) \] \[ W \approx 5 \times 10^{-9} \cdot 4.52 \times 10^{10} \] \[ W \approx 226 \text{ J} \] --- ## **Final Answer** \[ \boxed{ W \approx 226 \text{ J} } \] --- ## **Summary** - **The work needed to transport a charge of \(5\,\text{nC}\) from infinity to \((0,0,-5)\) is approximately \(226\,\text{J}\).** - **We used the image theory for the conducting plane and treated the spherically symmetric charge distribution as a point charge at the origin.**