Question

help, please.

5.0 cm 10 nc 2. a. What is the net Electric field at the point marked P (at the empty corner of the rectangle). Give magnitude and direction with respect to the + x axis. ------- + b. What would be the force (a vector) on a +2.50 C charge placed at point P? 7.0 cm ---- -5.0 nC 5.0 nC

Answer 1

T 5.0cm lonc ----7A t *hone Chevige is given hu Zn 5.one... -5.one soln . The electric field at P due lore. the thone charge is given by . Er 6; = LE (2) Dora 7. = (x109) x[ 65 x 10-9) 1. tone (0-07)2] (= q. 18x103 N/C ; along o direction all Electric field at I due to to ne charge - E, = (9x109 )x (10X10092 1 09 VE 10.05)? i lors! = 3.6x109 NIC ; along Ä p direction! so the net electric field due to 5.0 ne and lone charge is given by - E = √6,² + E2 Tas E, and Eq are perpendicubay =(9.18 x103)2 + (3.64101)2 3.72 x104 NLC; directed along BP direction 1. Now the electric field due to 5.0 ne charge is Eg = (9x109) * (-5X10-9) . 1. = 9x109× (-5X10-9) 4. [d = (0.05)²+(0.07) - 37.4x10-3y - = 7.4x10-31 = -6.08x103 N/C along B direction Here the negative sign indicates that the field is along PB direction. so the net electric field at ip, e n = [(3.72 x109)-(6.08 x103) ] NIC CS Camscanne3. 11 X104 N/C; dineded dong BP cs Scanned WIE d2

. If the net dectric field makes an O with the 0 x-axis, then we can write - worn and a tano = 'EL = 9.18x103) . 0.255 :, (3.60X104) 3) 0 14.31° Post @ The force on a charge +2.50 Mc placed at f 1 = (The net electric field at p) (the charge at I) = [63-11X10") x (2.50 x10-6)7 = 0.0777 N POP 7.77 810-2 N along B È direction ie, the force will also makes an anglen 14.310 with mo: the x-axis. o 117 E s 'Scanned with CamScanner , ...)
In this case the charge at P experiences the force along BP direction because the charge at P is positive. If the charge at P was negative then it would experiences the force along PB direction.