# Explanation based on the superposition principle the

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Explanation: Based on the superposition principle, the potential at the center due to the charges at the corners is V = V 1 + V 2 + V 3 + V 4 = k q r ( 1 1 + 1 + 1) = 0 , where r is the common distance from the center to the corners, so the work required to bring the charge q from infinity to the center is then W = q V = 0.
006 10.0points The distance between the K + and Cl ions in KCl is 2 . 8 × 10 10 m. Find the energy required to separate the two ions to an infinite distance apart, assum- ing them to be point charges initially at rest.
rguig (tr7356) – Homework 05 – yao – (59020) 3 1. 5 . 14 eV correct 2. 4 . 37 eV 3. 6 . 77 eV 4. 709 . 1 eV 5. 101 . 3 eV 6. 8 . 91 eV Explanation: Let : q = 1 . 6 × 10 19 C , k = 8 . 99 × 10 9 N · m 2 / C 2 , and d = 2 . 8 × 10 10 m . The energy is W = Δ K + Δ U = 0 U i = k ( e ) e d = k q 2 d = (8 . 99 × 10 9 N · m 2 / C 2 ) 2 . 8 × 10 10 m × (1 . 6 × 10 19 C) 2 parenleftbigg 1 eV 1 . 6 × 10 19 J parenrightbigg = 5 . 14 eV . 007 10.0points A dipole field pattern is shown in the figure. Consider various relationships between the electric potential at different points given in the figure. J P K Z L + Consider five potential relationships a) V P = V Z > V K b) V P = V Z = V K c) V P = V Z < V K d) V J < V K < V L e) V J > V K > V L Which relationships are correct? 1. ( a ) and ( e ) only 2. ( c ) and ( d ) only 3. ( d ) only 4. ( a ) and ( d ) only 5. ( b ) and ( e ) only 6. ( e ) only 7. ( a ) only 8. ( b ) and ( d ) only correct 9. ( c ) only 10. ( c ) and ( e ) only Explanation: The electric potential due to one single point charge at a distance r from the charge is given by V = k q r . For a dipole system, the total potential at any point is the sum of potentials due to one positive point charge and one negative point charge (Superposition Principle). From symmetry considerations, it is easy to see that the electric field lines are perpen- dicular to a line which passes through the midpoint K and points P and Z . No work needs to be done to move a positive test charge along the midplane because the force and the displacement are perpendicular to each other. V P = V K = V Z , relation ( b ). Furthermore, moving along the direction of a electric field line ( i.e. , moving in the direc- tion from positive charge to negative charge
rguig (tr7356) – Homework 05 – yao – (59020) 4 along the electric field line) always lowers the electric potential, because the electric field will do positive work to a positive test charge in order to lower its electric potential energy. Therefore, V J < V P by considering the line going from P to J , and V Z < V L by consider- ing the line going from L to Z . V J < V K < V L , relation ( d ). The correct choices are ( b ) and ( d ) only. 008(part1of2)10.0points Consider two “ solid ” conducting spheres with radii r 1 = 5 R and r 2 = 8 R , separated by a large distance so that the field and the poten- tial at the surface of sphere #1 only depends on the charge on #1 and the corresponding