1) Three equal charges of amount Q are placed at the vertices of a square of sid
ID: 1348631 • Letter: 1
Question
1) Three equal charges of amount Q are placed at the vertices of a square of side a as illustrated at the right. Determine: a) the total electric field (both magnitude and direction) 0 20 produced by the three charges at point A. Make sure to draw electric field vectors and its components on the figure and the Final E-vector and direction (4 pts) b) the total electric potential produced by the three c) the Coulomb force at point A (both magnitude and d) given the value of 0- 8.50 nC, a is 6.50 cm, find Q charges at point A direction) if there is a charge -0 placed at A numerical values of electric field and force.Explanation / Answer
a. Formula for electric field due to each charge is E = KQ/r^2
so E due to x components is
Ex = -k(2Q)/a^2 - kQ/2a^2
Ex = -kQ/a^2 (2 + 1/2)
Ex = -2.5 k Q/a^2
Similarly E due to all Y components is Ey = k(Q)/a^2 - kQ/2a^2
Ey = kQ/a^2 (1 - 1/2)
Ey = 0.5 k Q/a^2
Enet^2 = Ex^2 + Ey^2
E = sqrt(Ex^2 + Ey^2)
E = sqrt((-2.5 k Q/a^2)^2 + (0.5 k Q/a^2)^2)
E = 6.5 k Q/a^2
direction: 180 - arctan(0.5/2.5) = 169 degrees
b) Electric potential Due to
V = k (2Q)/a - k Q/a + k Q/(sqrt(2) a)
V = (kQ/a) (2 - 1 + 1/sqrt(2))
V = 1.707 k Q/a
c) magnitude of force F = Q E
F = 6.5 k Q^2/a^2
direction- 180 + 169
Theta = 349 degrees
d. magnitude of Electric field E = 6.5 k Q/a^2
E = 6.5*9e9*8.5e-9/(0.065*0.065)
E = 1.17 x 10^5 N/C
magnetude of force F = 6.5 k Q^2/a^2
F = 6.5*9e9*8.5e-9*8.5e-9/(0.065*0.065)
F = 1* 10^-3 N/C
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