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Unformatted text preview: Solution Derivations for Capa #4 1) Consider two separate systems, each with four charges of magnitude q arranged in a square of length L as shown above. Points a and c are in the center of their squares while points b and d are half way between the lower two charges. (Give ALL correct answers, i.e., B, AC, BCD...) QUESTION: A) The electric potential at d is zero. B) The electric potential at c is zero. C) The electric field at d is zero. D) The electric field at c is zero. E) The electric potential at b is zero. F) The electric field at a is zero. G) The electric field at b is zero. H) The electric potential at a is zero. ANSWER: A) False, the potential from the positive charges is greater than the potential from the negative charges since the positive charges are closer. B) True, it lies on the eqipotential line. C) False, the field from the positive charge cancels out, but there is still a field from the negative charges. D) False, the diagonal lines from q to c to + q both point toward q , resulting in a net field. E) True, the positive and negative charges nearest b cancel out as well as the pair furthers from it. F) True, see (E) G) False, there is a flux toward the lower, q . The upper two provide a lesser field that cancels out part of the field moving along the bottom from right to left. 1 H) True, a is equal distance from each point charge. As such, the potential sums to zero. Remember that potential is a scalar and electric field is a vector. CAPA is also looking for the letters of the correct answer, in this case, BEFH 2) Using the diagram above, find the magnitude of the electric field at point d . DATA: q = 0 . 750 μC , L = 0 . 40 m . q = Given L = Given The magnitude of the electric field is E = kq r 2 . Since the field due to the positive point charges will cancel each other and the horizontal components of the negative point charges will also cancel, only the vertical component of each negative point charge contributes to the field. The field due to the vertical component will be E = kq r 2 cos θ . Since there are two negative charges, the electric field is twice that from one charge, E = 2 kq r 2 cos θ r is found from the right triangle shown, and cos θ can also be calculated from the picture. Thus, E = 2 kq L 2 + L 2 2 L r L 2 + L 2 2 = 2 kqL L 2 + L 2 2 3 2 3) The figure below shows two points in an Efield: Point 1 is at ( X 1 , Y 1 ) = (3 , 4) in m , and Point 2 is at ( X 2 , Y 2 ) = (12 , 9) in m . The Electric Field is constant, with a magnitude 2 of 80 V/m , and is directed parallel to the +Xaxis. The potential at point 1 is 1000 V ....
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This homework help was uploaded on 04/03/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.
 Fall '08
 ROGERS
 Physics, Charge

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