x PA Electric Field of... x c Chegg Study Guided S x NDs NDSU North Dakota State
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x PA Electric Field of... x c Chegg Study Guided S x NDs NDSU North Dakota State... X Bb Assignments Phys 212 3 triton physi dak,ed uploaded%2fndsu%2f2T95501f1948a5625ndsu11%2fdefaull c a search B Messages Courses Help Logout Makenzie Reutter v Student section: 01 Physics 212 College Physics II Spring 2016 Main Menu Contents Grades Syllabus Timer Notes Evaluate Feedback Print a Info Course Contents Homework 02 Electric Field of Two Particles 92 91 In the figure, particle 1 of charge q 7.20X 10 C and particle 2 of charge q2 2.88X 10 C are fixed to an x axis, separated by a distance d 0.100 m. electric field E as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to Calculate their net the left. What is E(-0.100 Submit Answer Tries 0/5 What is E(-0.030 Submit Answer Tries 0/5 What is E(0.080 Submit Answer Tries 0/5 What is E(0.110)? Submit Answer Tries 0/5 NETGEAR20 Internet access ew Chronological View other Views MIvy general preferences on what is 7:33 PM 1/20/2016Explanation / Answer
Sol.`;- E(x)= -k [(q1/x^2)+(q2/(x-d)^2]
Now we can solve
E (-0.100) = -9 x 109 [(7.2 x 10-5/(-0.1)^2)+(2.88 x 10-4/(-0.1-0.1)^2]
Hence E (-0.100) = -12.96 x 10^7 N/C...................Ans.
Similarly
E (-0.030) = -9 x 109 [(7.2 x 10-5/(-0.03)^2)+(2.88 x 10-4/(-0.03-0.1)^2]
Hence E (-0.030) = -87.337 x 10^7 N/C...................Ans.
E (0.080) = -9 x 109 [(7.2 x 10-5/(0.08)^2)+(2.88 x 10-4/(0.08-0.1)^2]
Hence E (0.080) = -11.687 x 10^7 N/C...................Ans.
E (0.110) = -9 x 109 [(7.2 x 10-5/(0.11)^2)+(2.88 x 10-4/(0.11-0.1)^2]
Hence E (0.110) = -288.595 x 10^7 N/C...................Ans.
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