Two charges of -2.1 micro-Coulombs and a charge of +2.1 mC are placed on the cor

Question

Two charges of -2.1 micro-Coulombs and a charge of +2.1 mC are placed on the corners of an equilateral triangle where the length of a side is 44cm. what is the magnitude of the total electric field in the middle of the bottom side of the square? Answer in MN/C (1x10^6) -q 44cm +q Px -q y | |- - - x Px=origin r= .44m/2= .22m at Px q1= -2.1mC q2= -2.1mC q3= 2.1mC K= 8.99x10^9 MN^2/C^2 Do I use Coulombs Law? The + and - charges cause attraction +q-------> <--------- -q. -q and -q repel. .44/.44 will equal a 45 degree angle? so,F31= K|(2.1x10^-6)(2.1x10^-6)|/(.44)squared. which equals .205N F31x=.205N sin45 and F31y=.205N cos45 added together equal .145 +.145=.290N F21x=-.290N Answer is Zero?

Explanation / Answer

post the diagram so that i can judge.. howeva ua approach seems to be right... plss rate

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