what is the strength of the electric field at the position indicated by the dot in (figure 1)?
How can we determine the field’s direction in the area that is indicated by the dot in (Figure 1.)?
Answer:
Because the charges are equally far apart and both positive, they repel in the same amount. If you draw forces, it will be the case that they cancel the component y of the field that is final. There will be only the component x. Since they are identical, it will result in twice as large as the components of the x field.
E=kq/r^2
R= 5 + root2 = 7.07 cm or .0707 millimeters (property of 45-45-90 right-angled right triangle)
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
Then, multiply it by 2 to accommodate both charges.
E=2546.35 N/C, in the direction of 0 ° (no y component because they cancel one another out)
What is The Strength of The Electric Field At The Position Indicated by The Dot in (Figure 1)?
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