Question

MAC-e46520c3f8s2a54 Net Secure htmircourseld-1509863 Hw 2:Ch 20 4-7 1011 Problem 20.30- Enhanced-with Feedback What is the strength of the electric field at the position indicated by the dot in (Figure 1)? Express your answer in newtons per coulomb. Review 1 Constants Periodic Table N/C Figure Submit 1.0 nC ▼PartB 5.0 cm What is the direction of the electric field at the position indicated by the dot in the figure? Specity the direction as angle above the horizontal line. 5.0 cm Express your answer in degrees 5.0 cm Pearson

Answer 1

R = squareroot s^2 + 5^2 = 7.07 cm = 7.07 times 10^-2 m phi = tan^-1(5/2) = 45 degrees q = -1 nc Net electric field at dot E_Net = squareroot E_x^2 + E_y^2 where, E_x = horizontal component of electric field = Ecos phi - Ecos phi = -2Esin phi = -2kq/r^2sin phi \r\n = -29 times 10^9 times 1 times 10^-9 C/(7.07 times 10^-2 m)^2sin 45 degrees = -2546.35N/c because E_Net = squareroot 0^2 + (-2546.35)^2 E_Net = 2546.35 N/C Answer b) Direction theta = tan^-1(E_y/E_x) = tan 6-1(2546.35/0) theta = 90 degrees theta = 90 degrees clockwise from + x - axis ---Answer (Nagation sign indicate)