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What is The Strength of The Electric Field At The Position Indicated by The Dot in (Figure 1)?

    What is the direction of the electric field at the position indicated by the dot in (Figure 1)?

    How can we determine the field’s direction at the location shown by the dot (Figure 1.)?

    Answer:

    Because the charges are located in the same distance and both positive, they repel at the same rate. If you draw the forces they will cancel the part that is y in the field that is final. It will only be the component x. Since they are identical, it will result in twice as large as the component x in one electric field.

    E=kq/r^2

    R= 5 x root2 equals 7.07 cm , or .0707 millimeters (property of right triangle 45-45-90)

    E=(9 x 10^9)(1 x 10^-9 )/(.0707)^2

    = 1800.5 N/C

    To find the x component

    cos 45 = x/1800.5

    x= 1273.18 N/C

    Simply multiply the number by 2 to accommodate each charge,

    E=2546.35 N/C with a direction of 0 ° (no y component because they cancel one another out)

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