Positive point-charges of +16.0 mC are fixed at two of the vertices of an equilateral triangle with sides of 1.30 m, located in vacuum. Determine the magnitude of the E-field at the third vertex.(adsbygoogle = window.adsbygoogle || []).push({});

so...all the particals are in equalibrum so that means F=ma=0

so if i call the third vertex c...Ea+Eb+Ec=0 so that means that Ea+Eb=-Ec

now i can write kqa/r^2+kqb/r^2=Ec and i just have to solve for Ec right?

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# Charges on the vertex of a triangle

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