![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjf78Y96WuT57Kx__dlHPGkw2YVb5itDeP7F5Txt0oNsawwY6URKXcR2yfXNqt7ZWvVq6WwbiIJ0wX9W0A12bNlVAxM6HugTyASHUBWABAfmV-BkDyK1arFtgXjQN3GLYYmCre-cGmxUS_A/s800/GR0177%20%2360a.jpg)
An infinite, uniformly charged sheet with surface charged density σ cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in figure. The electric flux Φ through the Gaussian surface is:
A. πR²σ / ε0
B. 2πR²σ / ε0
C. π(R − x)²σ / ε0
D. π(R² − x²)σ / ε0
E. 2π(R² − x²)σ / ε0
(GR0177 #60)
Solution:![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj4UEeDjMIoFgg9Vo6wsoAUXMyhVenuFv1mQVgOy8Lo2WYjXtTxDdeQv6FbbFk7jnISqFsXxu4-jQyTG3h907co2pJbScZWQ3yzAvNEAR_JEFInvcf3SL_Tf8W9FP4KxXC3THfLBTj9SZ0X/s800/GR0177%20%2360b.jpg)
The Gaussian surface marks a circle around the plane with radius:
r² = R² − x²
The net charge within the Gaussian surface:
qenc = σA = σπr² = σπ(R² − x²)
Electric Flux:
Φ = qenc / ε0 = σπ(R² − x²) / ε0
Answer: D
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