Question

7- A uniform charge of +Q is placed on the sphere of a
Van de Graaff generator, such that electron discharge is observed from a conducting point a distance of 46 cm away ( center to center). Assume the dielectric strength of air, E = 3x10^6 N/C
(a) Determine the electric potentials at the 100 cm and 80 cm radial location
(b) A small air balloon of charge q=-8.4nC mass 1.52 grams, is placed at rest at the 100 cm potential surface, and then released. Determine its speed v at the 80cm radial location ( neglect any air resistance and work = ΔK)

Answer 1

Answer on Question#55046 - Physics - Electromagnetism

A uniform charge of $+Q$ is placed on the sphere of a Van de Graaff generator, such that electron discharge is observed from a conducting point a distance of 46 cm away (center to center). Assume the dielectric strength of air, $E = 3 \times 10^{6} \frac{\mathrm{N}}{\mathrm{C}}$

(a) Determine the electric potentials at the 100 cm and 80 cm radial location

(b) A small air balloon of charge $q = -8.4\mathrm{nC}$ mass $m = 1.52$ grams, is placed at rest at the 100 cm potential surface, and then released. Determine its speed $v$ at the 80 cm radial location (neglect any air resistance and work = ΔK)

Solution:

The electric field at the distance $r$ (greater than its radius) from the center of the charged sphere is given by

$$E = k_{e} \frac{Q}{r^{2}},$$

where $k_{e} = 8.988 \times 10^{9} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}}$ – is the electrostatic constant. Since for $r = 46\mathrm{cm}$ the electric field is given $E = 3 \times 10^{6} \frac{\mathrm{N}}{\mathrm{C}}$, we can find the charge of the sphere

$$Q = \frac{E \cdot r^{2}}{k_{e}} = \frac{3 \times 10^{6} \frac{\mathrm{N}}{\mathrm{C}} \cdot (0.46\mathrm{m})^{2}}{8.988 \times 10^{9} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}}} = 0.7\mu\mathrm{C}$$

(a) The potential of the charged sphere at the distance $r$ is given by

$$\varphi = k_{e} \frac{Q}{r}$$

Therefore the potential at $r = 100\mathrm{cm}$ is given by

$$\varphi(100\mathrm{cm}) = 8.988 \times 10^{9} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}} \frac{0.7\mu\mathrm{C}}{1\mathrm{m}} = 6292\mathrm{V}$$

The potential at $r = 80\mathrm{cm}$ is given by

$$\varphi(80\mathrm{cm}) = 8.988 \times 10^{9} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}} \frac{0.7\mu\mathrm{C}}{0.8\mathrm{m}} = 7865\mathrm{V}$$

(b) According to the law of conservation of energy, the difference in kinetic energies $E_{k}$ at these distances is equal to the difference in the potential energies $V$ at these distances of the balloon:

$$\begin{array}{l} E_{k}(80\mathrm{cm}) - E_{k}(100\mathrm{cm}) = V(100\mathrm{cm}) - V(80\mathrm{cm}) \\ E_{k}(80\mathrm{cm}) - E_{k}(100\mathrm{cm}) = q \cdot \left(\varphi(100\mathrm{cm}) - \varphi(80\mathrm{cm})\right) = \\ = -8.4\mathrm{nC} \cdot (6292\mathrm{V} - 7865\mathrm{V}) = 13.2\mu\mathrm{J} \end{array}$$

Since $E_{k}(100\mathrm{cm}) = 0$ and $E_{k}(80\mathrm{cm}) = \frac{m \cdot v^{2}}{2}$, we obtain

$$\frac{m \cdot v^{2}}{2} = 13.2\mu\mathrm{J}$$ $$v = \sqrt {\frac {2 \cdot 13.2\,\mu J}{m}} = \sqrt {\frac {26.4\,\mu J}{1.52\,\text{g}}} = 0.13\,\frac{\text{m}}{\text{s}}$$

**Answer:**

(a) $\varphi(100\,\text{cm}) = 6292\,\text{V}$

$\varphi(80\,\text{cm}) = 7865\,\text{V}$

(b) $0.13\,\frac{\text{m}}{\text{s}}$

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