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An electric field in a region is E = +200 i N/C for x > 0 and E = -200 i N/C for x < 0. A closed cylinder of length 2 m and cross-sectional area 1 m² has its axis along the x-axis with its centre at the origin. (Take epsilon0 = 8.85 x 10⁻¹² C² m⁻² N⁻¹.) Find the total charge enclosed by the cylinder.
- 3.54 x 10⁻⁹ C
- 0
- 1.77 x 10⁻¹¹ C
- 8.85 x 10⁻¹² C
Correct answer: 3.54 x 10⁻⁹ C
Solution
The field points outward through both end caps (out the +x cap where E is +x, and out the -x cap where E is -x), each contributing E*A. The curved surface has E parallel to it, so zero flux there. Net flux = 2*E*A = 2*200*1 = 400 N m²/C. Enclosed charge q = epsilon0 * flux = 8.85e-12 * 400 = 3.54e-9 C. (The printed area '10² m²' is a typo for 1 m²; with A = 1 the standard answer is 3.54 nC.)
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