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point charge 1.5*10to the power -3 and 0.5 are placed at corner A and C of a right angled triangle ABC such that AB= 1.2m and BC= 0.5m find the resultant electric field at B
Asked by shashinayaka032 | 01 Dec, 2023, 08:52: PM
Electric field EA at B due to 1.5 × 10-3 C charge at A is

EA = K × ( 1.5 × 10-3 ) / (1.2)2  N/C

Electric field EC at B due to 0.5 × 10-3 C charge at C is

EC = K × ( 0.5 × 10-3 ) / (0.5)2  N/C

where K = 9 × 109 N m2 C-2 is Coulomb's constant.

EA = 9 × 109 × ( 1.5 × 10-3 ) / (1.2)2  N/C = 9.375 × 106 N/C

EC = 9 × 109 × ( 0.5 × 10-3 ) / (0.5)2  N/C = 18 × 106 N/C

Resultant electric field  = [ (9.375)2 + (18)2 ]1/2 * 106 N/C = 2.03 × 107 N/C

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