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**Unformatted text preview: **1 1. The point charge Q below is stationary. The electric potential energy U of the system is- 37 nJ if point charge q is at A . As q moves from A to B , the work done on q (by the electric field of Q ) is +23 nJ. What is U when q is at B ? [2] Q × × A B q Δ U = U B- U A =- (Work done by vector E Q ) =- 23 nJ U B = U A + Δ U = (- 37 nJ) + (- 23 nJ) =- 60 nJ 2. When the capacitors below are fully charged, the 6.00 μ F-capacitor has a charge of 36.0 μ C. What is the charge of 4.00 μ F-capacitor? [2] 3.00 μ F 6.00 μ F 4.00 μ F Q 3 . 00 μ F = 3 . 00 μ F 6 . 00 μ F Q 6 . 00 μ F = 18 . μ C Q 4 . 00 μ F = Q 3 . 00 μ F + Q 6 . 00 μ F = (18 . μ C) + (36 . μ C) = 54 . μ C 2 3. In the diagram below, q = 0 . 10 nC and d = 3 . 0 cm. Find the x-component E x of the net electric field at the point with coordinates (- 4.0 cm, +4.0 cm). [4] r + q = √ 17 . 0 cm r- q = √ 65 . 0 cm cos θ + q = 1 . √ 17 . cos θ- q = 7 . √ 65 . x y- q + q d d × θ + q θ- q vector E + q vector E- q E x =-| vector E + q | cos θ + q + | vector E- q | cos θ- q =- 8 . 2 N/C 4. A point charge q = 2 . 40 nC is at the center of a spherical conducting shell (inner radius of 1.00 cm, outer radius of 1.20 cm). The electric field 4.00 cm from the center of the shell has a magnitude of 5.00 kV/m and points radially inward. Find the surface charge density on the outer surface of the shell. (Justify your choice of sign.) [3] q | Q outside | = 4 πǫ (4 . 00 cm) 2 | vector E | = 8 . 89 × 10- 10 C | σ | = | Q outside | 4 π (1 . 20 cm) 2 = 4 . 91 × 10- 7 C/m 2 σ < 0 : vector E inward 3 5. Three long straight parallel wires (seen end-on below) carry currents out of the page; a = 2 . 50 cm. Find the magnitude of the50 cm....

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