# Capacitance of an Isolated Sphere ## What is Capacitance of a Sphere? - **Capacitance** of a charged sphere is defined as the **charge per unit potential at the surface of the sphere** - This relationship is expressed as: $$C = \frac{Q}{V}$$ - Where: - $C$ = capacitance (F) - $Q$ = charge (C) - $V$ = potential difference (V)

Capacitance of a sphere: C = Q / V, where Q is the charge stored on the surface and V is the potential at the surface.

## The Spherical Conductor Approximation - The charge distributed on the **surface** of a spherical conductor can be modelled as a **point charge at its centre** - This simplification is valid when considering points outside the sphere - The potential $V$ of an isolated point charge is given by: $$V = \frac{Q}{4\pi\varepsilon_0 R}$$ - Where: - $R$ = radius of sphere (m) - $\varepsilon_0$ = permittivity of free space (F m$^{-1}$) - **Important**: The charge $Q$ is not the charge of a capacitor plate, but the **charge stored on the surface of the spherical conductor** ## Deriving the Capacitance Formula - We can combine the two equations for capacitance and potential to derive an expression specifically for spheres - Starting with both equations: $$V = \frac{Q}{4\pi\varepsilon_0 R}$$ $$V = \frac{Q}{C}$$ - Since both expressions equal $V$, we can equate them: $$\frac{Q}{4\pi\varepsilon_0 R} = \frac{Q}{C}$$ - Rearranging gives the **capacitance of an isolated sphere**:

The capacitance of an isolated sphere is directly proportional to its radius: C = 4πϵ₀R.

$$C = 4\pi\varepsilon_0 R$$ ## Key Properties | Property | Relationship | |----------|-------------| | Capacitance vs. radius | $C \propto R$ (directly proportional) | | Units | Farads (F) | | Depends on | Radius and permittivity only | | Independent of | Charge or potential | - Larger spheres have **greater capacitance** for the same charge - The capacitance depends only on the **geometry** (radius) and the **medium** (permittivity) - Doubling the radius doubles the capacitance ### Visualizing Capacitance vs Radius The graph below illustrates the linear relationship between capacitance and radius for an isolated conducting sphere:

- The straight line through the origin confirms the direct proportionality: $C = 4\pi\varepsilon_0 R$ - The gradient of the line equals $4\pi\varepsilon_0 = 1.11 \times 10^{-10}$ F m$^{-1}$ ## Worked Example A conducting sphere used in a physics demonstration has a radius of 25 cm. It is charged to a potential of 750 kV. After an electrical discharge through the air, the sphere retains only 8% of its original energy. Calculate: a) The capacitance of the sphere b) The potential of the sphere after the discharge ### Answer: **Part (a)** **Step 1: List the known quantities** - Radius of sphere, $R = 25$ cm $= 25 \times 10^{-2}$ m - Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12}$ F m$^{-1}$ **Step 2: Write the equation for capacitance of an isolated sphere** $$C = 4\pi\varepsilon_0 R$$ **Step 3: Calculate the capacitance** $$C = 4\pi \times (8.85 \times 10^{-12}) \times (25 \times 10^{-2})$$ $$C = 2.78 \times 10^{-11} \text{ F}$$ **Part (b)** **Step 1: List the known quantities** - Original potential, $V_1 = 750$ kV $= 750 \times 10^3$ V - Final potential = $V_2$ - Original energy = $E_1$ - Final energy, $E_2 = 0.08 E_1$ **Step 2: Write the equation for energy stored by a capacitor** $$E = \frac{1}{2}CV^2$$

Energy stored in a capacitor depends on V², not just V.

**Step 3: Write equations for energy before and after discharge** $$E_1 = \frac{1}{2}CV_1^2$$ $$E_2 = \frac{1}{2}CV_2^2$$ **Step 4: Equate the expressions and simplify** Since $E_2 = 0.08 E_1$: $$\frac{1}{2}CV_2^2 = 0.08 \left(\frac{1}{2}CV_1^2\right)$$ $$V_2^2 = 0.08(V_1^2)$$ $$V_2 = \sqrt{0.08} \times V_1$$ **Step 5: Calculate the final potential, $V_2$** $$V_2 = \sqrt{0.08} \times (750 \times 10^3)$$ $$V_2 = 2.12 \times 10^5 \text{ V} = 212 \text{ kV (3 s.f.)}$$