Charges of 4.0 muC and - 6.0muC are placed at two corners of an equilateral tria
ID: 1509931 • Letter: C
Question
Charges of 4.0 muC and - 6.0muC are placed at two corners of an equilateral triangle with sides of 0.10 m. At the third corner, what is the electric field created by these two charges? Two point charges each have a value of 30.0mC and are separated by a distance of 4.00 cm. What is the electric field midway between the two charges? A charge of +2 C is at the origin. When charge Q is placed at 2 m along the positive x axis, the electric field at 2 m along the negative x axis becomes zero. What is the value of Q? Two equal charges, each Q. are separated by some distance. What third charge would need to be placed half way between the two charges so that the net force on each charge would be zero? zero?Explanation / Answer
Electric field due to point charge = kq/r^2
So, Electric field at the third point = k[4[cos(60)i + sin(60)j] - 6[-cos(60)i + sin(60)j]]*10^-9/0.1^2 = 88900[10cos(60)i - 2sin(60)j] = 889000cos(60)i - 177800sin(60)j = 444500 i - 153979.3167j
|E| = sqroot(Ex^2 + Ey^2) = 470414.5831 V/m