Isosceles 48443

Three equal positive charges Q are located at the vertices of an isosceles right triangle ABC. The right angle is at vertex A. The length of side AB is 1m. What is the electric field strength at the center S of side BC, i.e., what force would act on a positive charge q of magnitude 1C?

Correct answer:

E =  53925315703.824 V/m

Step-by-step explanation:

c=1 m b=c=1 m ε=8.8541871012=8.854187 53925315703.824128.85421012 F/m  Q1=1 C Q2=Q1=1 C Q3=Q1=1 C  c2 + b2 = a2  a=2 c=2 1=2 m1.4142 m s=a/2=1.4142/20.7071 m AS=s=0.70710.7071 m AS=BS=CS=s  E1=4π ε1 s2Q1=4 3.1416 8.854210121 0.70712117975105234.608 V/m E2=E1=17975105234.608 V/m E3=E1=17975105234.608 V/m  E=E1+E2+E3=17975105234.608+17975105234.608+17975105234.608=53925315703.824 V/m5.3931010 V/m

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Tips for related online calculators
Our vector sum calculator can add two vectors given by their magnitudes and by included angle.
See also our right triangle calculator.
Calculation of an isosceles triangle.
See also our trigonometric triangle calculator.

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