4.) Given : Formula Q = 2 MC E = K Q R = 10 Cm - 0.10m ( R t r ) 2 Therefore , the magnitude of the solution : electric field abode the sphere E = K Q surface is 8. 0 x 105 NIC ( R tr ) 2 = (8. 985 x 10 " H. IZ ( 2 x 10 # ( ) DE = 7. 95 x 105 NIC C. 2 ( 0. 10 m +0.05 m) 2 = ( 8 . 908 7 10 " N m E = 8. 0 X 105 M / C ( 2 x10-6 ( ) C - 2 (0.15 m ) 2 = 17 976 N m2 ( 0. 02 25 m 24. A metal sphere of radius 10 cm carries a charge of + 2.0 microCoulomb uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point 5.0 cm outside the sphere surface? Given r = 10 cm = 0.1m E = q/0.A r=5cm = 0.05m( outside) E= 2 x10 -6 C / 8.988x 109 N.m2 C-2 (0.2826m2) q =+2 E =2 x10 6 C/ 4.01 x10 8 C A = 42 (area of sphere) E = 2 x 10-6 /2.54 x10 9N/C =42 E =7.87 x 10-16 N/C ( 5 points) A = 42 = 0.2826m2 (3 points) Find: E=? Solution EA= q/0 E = q/0.A