Point charge q1 = -5.05 nC is at the origin and point charge q2 = +2.90 nC is on

Question

Point charge q1 = -5.05 nC is at the origin and point charge q2 = +2.90 nC is on the x-axis at x = 2.80 cm. Point P is on the y-axis at y = 4.50 cm. (a) Calculate the electric fields E with arrow1 and E with arrow2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors. E with arrow1 = N/C i + N/C j E with arrow2 = N/C i + N/C j (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form. E with arrow = N/C i + N/C j Point charge q1 = -5.05 nC is at the origin and point charge q2 = +2.90 nC is on the x-axis at x = 2.80 cm. Point P is on the y-axis at y = 4.50 cm. (a) Calculate the electric fields E with arrow1 and E with arrow2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors. E with arrow1 = N/C i + N/C j E with arrow2 = N/C i + N/C j (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form. E with arrow = N/C i + N/C j Point charge q1 = -5.05 nC is at the origin and point charge q2 = +2.90 nC is on the x-axis at x = 2.80 cm. Point P is on the y-axis at y = 4.50 cm. (a) Calculate the electric fields E with arrow1 and E with arrow2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors. E with arrow1 = N/C i + N/C j E with arrow2 = N/C i + N/C j (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form. E with arrow = N/C i + N/C j

Explanation / Answer

a) E1 = k*q1/y^2

= 9*10^9*5.05*10^-9/0.045^2

= 22444 N/c

E2 = k*q2/(x^2 + y^2)

= 9*10^9*2.9*10^-9/(0.028^2 + 0.045^2)

= 9292 N/c

E1 = -22444 j N/C

E2 = -E2*cos(theta)i + E2*sin(theta)

= -9292*2.8/sqrt(2.8^2+4.5^2) i + 9292*4.5/sqrt(2.8^2+4.5^2) j

= (-4909 i + 7890 j) N/c

b) E = E1 + E2

= -22444 j -4909 i + 7890 j

= (4909 i -14554 j)N/C

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