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# hw6 - Physics 21 Fall 2009 Solution to HW-6 23-40 Two large...

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Physics 21 Fall, 2009 Solution to HW-6 23-40 Two large, parallel conducting plates carrying op- posite charges of equal magnitude are separated by d = 2 . 20 cm. (a) If the surface charge density for each plate has magnitude 47.0 nC/m 2 , what is the magnitude of E in the region betwixt the two plates? (b) What is the poten- tial difference between the two plates? (c) If the separation between the plates is doubled while the surface charge den- sity is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference? (a) Two parallel plates with opposite surface charge form a capacitor. We have discussed the field E between capacitor plates in terms of the field σ/ 2 0 of a single plate. From the equation sheet, the field between the plates is E = σ 0 = 47 . 0 × 10 - 9 Cm - 2 8 . 85 × 10 - 12 C 2 N - 1 m - 2 = 5310 N C (b) To find the potential difference, integrate along the sim- plest path between the plates: a straight line from the neg- ative to the positive plate and perpendicular to both. For this path, the potential will increase since the direction of integration is opposite to the field. V = E · dl = Ed = 5310 N C ( . 022 m) = 117 V (c) The electric field E doesn’t depend on the distance be- tween plates; it only depends on σ . For this problem, σ stays the same as d is increased. Therefore the electric field remains constant, and the potential difference Ed increases.

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hw6 - Physics 21 Fall 2009 Solution to HW-6 23-40 Two large...

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