Uniform Electric Field - General Physics - Solved Past Paper, Exams of Physics

Download Uniform Electric Field - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 2. (20 pts) A particle having a charge q = 3 × 10−9 C moves from point a to point b, separated by a distance of d = 0.5 m, along a straight line through a uniform electric field of magnitude E = 200 N/C. The electric field points from point a to point b. a) How much force does the electric field exert on the charge? F = qE = ( 3 × 10−9 ) (200) = 6 × 10−7 = 600 nN b) How much work is done on the charge by the electric field during this motion? W = ~F · ~ℓ = ( 6 × 10−7 ) (0.5) = 3 × 10−7 = 300 nJ c) By how much does the electrical potential energy change during this motion? Be sure your answer indicates whether the change is an increase or decrease. W = −∆U = −300 nJ Note that because a positive charge is moving in the same direction as the electric field, the electric potential is decreasing and thus so is the electric potential energy of the charged object. d) Find the electric potential difference between points a and b. ∆V = ∆U q = −3 × 10−7 3 × 10−9 = −100 V Again, since we’re moving in the direction of the electric field, the electric potential should be decreas- ing, hence the negative sign above.