What is the potential energy at the center of an equilateral triangle if +5esu, +5esu and -10esu are placed at the corners of the triangle? The triangle has a side length of √3cm.(a) 20V(b) 5V(c) \(\frac {5}{3}\)V(d) 0VI have been asked this question in an international level competition.Enquiry is from Potential Energy of a System of Charges in portion Electrostatic Potential and Capacitance of Physics – Class 12

What is the potential energy at the center of an equilateral triangle

1.	What is the potential energy at the center of an equilateral triangle if +5esu, +5esu and -10esu are placed at the corners of the triangle? The triangle has a side length of √3cm.(a) 20V(b) 5V(c) \(\frac {5}{3}\)V(d) 0VI have been asked this question in an international level competition.Enquiry is from Potential Energy of a System of Charges in portion Electrostatic Potential and Capacitance of Physics – Class 12
Answer» The correct choice is (d) 0V The best I can EXPLAIN: Distance of the center of the triangle from EVERY corner point is = \(\FRAC {2}{3} * \frac {\SQRT 3}{2}\) * √3cm=1cm [as we know that the center divides the median in a 2:1 ratio and the length of the median is \(\frac {2}{3}\)*side length]. Therefore, the potential at the center is =\(\frac {5}{1} + \frac {5}{1} -\frac {10}{1}\)=0 esu Volt.

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