?A point charge q 1 = -6.00 nC is at the point x = 0.600 m, y = 0.800 m, and a s

Question

?A point charge q1 = -6.00 nC is at the point x = 0.600 m, y = 0.800 m, and a second point charge q2 = 5.00 nC is at the point x = 0.600 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

A point charge q1 =-6.00 nC is at the point x = 0.600 m, y = 0.800 m, and a second point charge q2 = 5.00 nC is at the point x = 0.600 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges N/C o (counterclockwise from the +x-axis is positive) (counterclockwise from the x-axis is positive)

Explanation / Answer

here,

electric feild due to q1 , E1 = k * q1/(0.6^2 + 0.8^2)

E1= 9 * 10^9 * 6 * 10^-9 /( 0.6^2 + 0.8^2)

E1 = 54 N/C

theta = arctan(0.8/0.6)

theta = 53.13 degree

E1 = 54 * ( cos(theta) i + sin(theta) j)

E1 = (32.4 i + 43.2 j )N/C

electric feild due to charge q2 , E2 = k * q2/(0.6)^2

E2 = 9 * 10^9 * 5 * 10^-9 /( 0.6^2)

E2 = - 125 i N/C

net electric feild at origin , E = E1 + E2

E = - (92.6 i + 43.2 j ) N/C

|E| = sqrt(92.6^2 + 43.2^2)

|E| = 102.18 N/C

theta' = arctan(43.2/92.6)

theta' = 155 degree

the net electric feild at the centre is 102.18 N/C and 155 degree counterclockwise from +x axis

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