4. Three point charges, A (+2.0 JC), B (+4.0 JC) and C (-6.0 JC) sit consecutively in a line. If A and B are separated by 1.0 m and B and C are separated by 1.0 m, what is the net force on each charge? [7 marks] A I m Im For A, Fnet=FAB[left]+FAc[right] Fnet=(((kqAqB)/TAB?)-((kqAqc)/rAc)) [left] Fnet=((((9.00x109 N.m'/C2)(2.0x10-C)(4.0x10-C))/(1.0m)2) - (((9.00x109 N.m2/C2)(2.0x10-C)(6.0x10-C))/(2.0m)?)) [left] Fnet=0.045N [left] So, the net force of A is 0.045N to the left. For B, Fnet= FBA[right]+FBC[right] Fnet=((((9.00x109 N.me/C2) (4.0x10-C) (2.0x10-C))/(1.0m)?) + (((9.00x109 N.m2/C2)( 4.0x10-C)(6.0x10-C))/(1.0m)?)) [right] Fnet=0.288N [right] So, the net force of B is 0.288N to the right. For C, Fnet= FCA [left]+ FCB [left] Fnet=((((9.00x109 N.m2/C2) (6.0x10-C) (2.0x10-6C))/(2.0m)?) + (((9.00x109 N.m2/C2)(6.0x10-C)( 4.0x10-C))/(1.0m)?)) [left] Fnet=0.243N [left]