Point charge q 1 = -5.05 nC is at the origin and point charge q 2 = +3.15 nC is

Question

Point charge q1 = -5.05 nC is at the origin and point charge q2 = +3.15 nC is on the x-axis at x = 2.85 cm. Point P is on the y-axis at y = 4.10 cm.

1 = N/C + N/C 2 = N/C + N/C Point charge q1 = -5.05 nC is at the origin and point charge q2 = +3.15 nC is on the x-axis at x = 2.85 cm. Point P is on the y-axis at y = 4.10 cm. Calculate the electric fields at point P due to the charges q1 and q2. Express your results in terms of unit vectors. Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Explanation / Answer

theta = tan^-1(4.1/2.85) = 55.2 degrees


E1 = k*q1/0.041^2 = 9*10^9*5.05*10^-9/0.041^2 = 2.7*10^4 N/c

E1 = 0 i - 2.7*10^4 j

E2 = k*q2/(0.0285^2+0.041^2)

   = 9*10^9*3.15*10^-9/(0.0285^2+0.041^2)

    = 1.137*10^4 N/c

E2 = -E2*cos(55.2) i + E2*sin(55.2) j

= -6.489*10^3 i + 9.34*10^3



E = (E1x+E2x) i + (E1y+E2y) j

= -6.489*10^3 i + 36.34*10^3 N/c

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