# Three charges are placed at the corners of a 45-45-90 degree triangle. Let q45 = +3.2*10^-19 C, q45 = +3.2*10^-19 C and q90 = +6.4*10^-19 C. If the legs of the triangle are of length 1 meter,...

Three charges are placed at the corners of a 45-45-90 degree triangle. Let q45 = +3.2*10^-19 C, q45 = +3.2*10^-19 C and q90 = +6.4*10^-19 C. If the legs of the triangle are of length 1 meter, find the magnitude of the electric field at point P, equidistant from each charge.

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### 1 Answer

Three charges are placed at the corners of a 45-45-90 degree triangle. The charges are q45 = +3.2*10^-19 C, q45 = +3.2*10^-19 C and q90 = +6.4*10^-19 C. The legs of the triangle are of length 1 meter.

The point equidistant from each of the vertices is the circumcenter. For any triangle, the circumradius is given by sqrt((a^2+b^2+c^2)/(8*(1+cos A*cos B*cos C))). For the triangle in the problem the circumradius is sqrt(3/8)

The magnitude of the electric field due to a charge Q at a distance r from it is given by E = k*Q/r^2 where k = 9*10^9

For the three charges, the field at the point equidistant from them is teh equal to the sum of the field due to each charge. E = 9*10^9*(3.2*10^-19+3.2*10^-19+6.4*10^-19)/(3/8)

=> 3.072*10^-8 N/C

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