Question

= or Find dg in de 1cm 1-52 2-The electric field along the perpendicular bisector of a dipole of size Imm, at a distance of 2cm from the center of the dipole, is 8000 N/C. The charge of the dipole is doubled, and so is the size of the dipole. What is the field now? (Show all your work.) SOOC 9x109 (2)(916103103) Coro2 = 5000 (2.0*102)

Answer 1

at point P Electric field aloy perpendicular bisutor trgiven of the 25,4+) es Et PRE=ETE E a 2kq E=E+E 2 y 2)²/ 2 E = doooNlc 4 (x+2)/2 x = perpendicular distance tq- from the centre of ar dipole to point p. - 2x lomm' na distance from the center to a Length of clipele = 2r. z 1x10 m podatere For x777 r we have the approximation for Eis, E- 2kqor x 2000 - (9x109) (1X163) (9) (2x103) 3

q = 2000 x 8x lob - 8x8 9x109x Tato9 -1109 q = 7. 11 X 159 C -371Lnc. Now, q² = 2q = 2 *7.11 x 15°C _ q?___E 14.222 C. Size of dipoles 2(27) S ex (1X 10?). 2016 mi E = 2(2k q r = 2E E'E 2 X 7.11 nc LE'. 14:22 nc