What is the total charge enclosed within the sphere r < 2 cm, if the electric field intensity inside the sphere is given by E = f V/m, where %3D Po = 6 x 10-9 O a. 1.51 pC Ob. 3.02 pC O c 0.90 pC Od. 2.26 pC Oe. 1.36 pC 1.0.30 pC
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A: Electric flux φ = E̅•A̅ = EA Cosθ
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Q: L. ....! 10.86 cm 21 cm 31 cm
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A: Given: The electric field is 8x1013 N/C. The cross-section area is 4.1x10-3 m2. The angle is 60o.
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Q: IS a ge. Wld ough the sur the spner units of N m2/C? Select one: O a. 1.75 x105 O b. 1.7x 10-16 O .…
A: Use Gauss's Law to find flux Flux=sum of all charges within the surfaceε0 value of ε0=8.85×10-12…
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A: Using vector form of electric field at a point we can get the answer.
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Q: Find the electric flux ( in units of N.m-/C) through the surface in the figure (dashed line) given…
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A: We will answer the question using Gauss's law. The details are as follows.
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Q: A solid insulating sphere with radius R carries electric charge which is non-uniformly distributed…
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Q: 3. A long conducting metal cylinder of radius a is surrounded by a conducting metal thin-shelled…
A: a. From Gauss law, the formula for the electric flux through Gaussian surface is: Also, the formula…
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A: Given : A = 7.5 cm2 = 7.5 × 10-4 m2 E = 9.3 × 105 N/C θ = 90° (because electric field is parallel…
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