Answer: pi R 2 E.
- A 0
- B pi R<sup>2</sup> E
- C 2 pi R<sup>2</sup> E
- D 4 pi R<sup>2</sup> E
Correct answer: B. pi R<sup>2</sup> E
Explanation: Flux through curved surface = flux through flat face = E x pi R<sup>2</sup> (by Gauss's law for closed hemisphere, net flux = 0, so curved = flat face flux = pi R<sup>2</sup> E).
Field lines (radial) and equipotential lines (concentric circles) of a point charge. Equipotential surfaces are always perpendicular to the field lines. Image: Sjlegg, Public Domain, via Wikimedia Commons.
Concept context
Electric potential, equipotential surfaces, potential energy of charge systems, conductors, dielectrics, capacitors, combinations, and energy storage.
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