72 91 In the figure, particle 1 of charge q, -8.00x10 C and particle 2 of charge

Question

72 91 In the figure, particle 1 of charge q, -8.00x10 C and particle 2 of charge q2 1.60x10 C are fixed to an x axis, separated by a distance d 0.100 m. Calculate their net electric field E(x) 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 the left. What is E(-0.100)? Sumt Answer Incorrect. Tries 3/5 Previous Tries What is E(-0.030)? 109387755NIC Submit Answer Tries 1/5 Previous Tries What is E(0.020) Submit Answer Tries 0/s What is E(0.130)? Submit Answer Tries o/s

Explanation / Answer

1)
x = - 0.1 m
E = - kq1/x2 - kq2 / (x+d)2
= - 8.99 x 109 x [8 x 10-6 / (0.1)2 + 1.6 x 10-5 / (0.2)2]
= - 7.23 x 106 N/C

2)
x = - 0.03 m
E = - kq1/x2 - kq2 / (x+d)2
= - 8.99 x 109 x [8 x 10-6 / (0.03)2 + 1.6 x 10-5 / (0.13)2]
= - 8.00 x 107 N/C

3)
x = 0.02 m
E = kq1/x2 - kq2 / (d - x)2
= 8.99 x 109 x [8 x 10-6 / (0.02)2 - 1.6 x 10-5 / (0.08)2]
= 1.80 x 108 N/C

3)
x = 0.13 m
E = kq1/x2 + kq2 / (d - x)2
= 8.99 x 109 x [8 x 10-6 / (0.13)2 + 1.6 x 10-5 / (- 0.03)2]
= 5.85 x 106 N/C

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