what is the strength of the electric field at the position indicated by the dot in (figure 1)?
![](https://secuesite.com/wp-content/uploads/2021/12/1111-1.jpg)
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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