A point particle with charge q = 4.2 C is placed on the x axis at x = 10 cm and

Question

A point particle with charge q = 4.2 C is placed on the x axis at x = 10 cm and a second particle of charge Q = 7.5 C is placed on the x axis at x = +25 cm.

(a) Determine the x and y components of the electric field due to this arrangement of charges at the point (x, y) = (10, 10) (the units here are centimeters).

(b) Determine the magnitude and direction of the electric field at this point.

The "(?)" are the answers I need. I have provided the units that I need the asnwers in on the third column of the table.

Explanation / Answer

E due to the charge at 4.2C at the point (x, y) = (10, 10)=(9X109X4.2X10e-6)/5 =7.56X10e-6

direction of this E at the point (x, y) = (10, 10)=tan-1(1/2)=26.560counter-clockwise from the +x axis

similarly

E due to the charge Q = 7.5 C is placed on the x axis at x = +25 cm=(9X109X7.5X10e-6)/3.25 =20.76X10e-6

direction of this E at the point (x, y) = (10, 10)=tan-1(2/3)=33.690counter-clockwise from the +x axis

so EX due te charge  4.2 C = 7.56X10e-6 cos 26.56=6.76X10e-6 N/C

EY due te charge  4.2 C = 7.56X10e-6 sin 26.56=3.38X10e-6 N/C

similarly

EX due te charge  7.5 C = 20.76X10e-6 cos 33.69=17.27X10e-6

EY due te charge  7.5 C = 20.76X10e-6 sin 33.69=11.51X10e-6

finally

ANSWERS

Ex due to two charges= add X components due to two charges.

Ey due  two charges= add Y components due to two charges.

MAGNITUDE=SQRT(Ex2+Ey2)

direction= tan-1(EY/EX)

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