Thank you for visiting In this week s lecture 3 Walter Lewin lit a fluorescent bulb by holding it near a Van de Graaff generator charged to 300 000. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
The potential difference between the two ends of the fluorescent tube, when held near a Van de Graaff generator is around 54,189 volts, calculated based on the dimensions and charge of the generator and the bulb.
Explanation:
The potential difference between the two ends of the bulb can be determined using the principles of physics and the specifics given in the question. As mentioned, the Van de Graaff generator is charged to 300,000 V. The potential of the charged conducting sphere is the same as that of an equal point charge at its center, so the entire 300,000 V is available at the surface of the sphere. However, only a portion of that voltage is experienced by the fluorescent bulb. The voltage drops linearly from the sphere's surface to the ground (which is at 0 V). Therefore, the potential difference the bulb experiences can be calculated as follows:
The sphere is 61.83 cm / 2 = 30.915 cm in radius. Therefore, the total distance from the sphere center to the ground via the bulb is 30.915 cm + 28.25 cm + 97.7 cm = 156.865 cm. The voltage at the near end of the bulb is (30.915 cm + 28.25 cm) / 156.865 cm * 300,000 V = 113,355 V. The voltage at the far end of the bulb is 30.915 cm / 156.865 cm * 300,000 V = 59,166 V. Therefore, the potential difference between the two ends of the bulb is 113,355 V - 59,166 V = 54,189 V.
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