Point charge q 1 = -5.30 nC is at the origin and point charge q 2 = +3.30 nC is

Question

Point charge q1 = -5.30 nC is at the origin and point charge q2 = +3.30 nC is on the x-axis at x = 2.75 cm. Point P is on the y-axis at y = 4.30 cm.

+ 1/6 points | Previous Answers YF13 21.P.034 point charge q1 =-5.30 nC is at the origin and point charge q2 = +3.30 nC is on the x-axis at x = 2.75 cm. Point pis on the y-axis at y = 4.30 cm. (a) Calculate the electric fields E1 and E2 at point P due to the charges q1 and q2. Express your results in terms of unit vectors. E, = 0 eids E, ant pint P due to the × N/C +10.84 N/Ci 0.84 2 Enter a number. (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form E=-1 N/Ci 0.3 × N/C Submit Answer Save Progress Practice Another Version

Explanation / Answer

electric field is given by

E = kQ/d^2

then electric field due to q1

E1 = 9*10^9*5.3*10^(-9)/(0.043)^2 = 25797.73 N/C

direction is -y axis

vector E1 = - 25797.73 j

then electric field due to q2

E2 = 9*10^9*3.3*10^(-9)/(0.043^2 + 0.0275^2) = 11400.05 N/C

now angle between E2 and +y axis

theta = arctan(2.75/4.3) = 32.6 degrees

now angle from +x axis = 90 + 32.6 = 122.6 degrees

vector E2 = 11400.05*cos(122.6) i + 11400.05*sin(122.6) j = - 6142 i + 9603 j N/C

net electric field E = E1 + E2 = - 6142 i + - 25797.73 j + 9603 j = (- 6142 i - 16194.73 j ) N/C

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