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![a \ student \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\ \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!](https://tex.z-dn.net/?f=%20a%20%5C%20student%20%20%5C%20is%20%5C%20able%20%5C%20to%20%5C%20lift%20%5C%20a%20%5C%20500%28n%29%20%5C%5C%20weight%20%5C%20by%20%5C%20applying%20%5C%20only%20%20%5C%20100%28n%29%20%5C%5C%20%5C%5C%20so%2C%20%5C%20this%20%5C%20info%20%5C%20here%2C%20%5C%20we%20%5C%20simply%20%5C%20divide%20%5C%20by%20%5C%5C%20how%20%5C%20much%20%5C%20this%20%5C%20kid%20%5C%20lifted%2C%20%5C%20by%20%5C%20the%20%5C%20weight%20%5C%20he%2Fshe%20%5C%5C%20%5C%20is%20%5C%20applying.%20%5C%5C%20%5C%5C%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B500%28n%29%2F100%28n%29%7D%7D%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%5C%5C%20and%20%5C%20from%20%5C%20this%2C%5C%20your%20%5C%20answer%20%5C%20would%20%5C%20%5C%5C%20conclude%20%5C%20to%20%5C%20be%20%5C%20%285%29%20%5C%5C%20%5C%5C%20%5Cboxed%7B5%7D%20%5C%20would%20%5C%20be%20%5C%20your%20%5C%20answer%21)
Hope this helps you!
Hope this helps you!
Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th
1 answer:
Answer:
E_total = 3 N / A
Explanation:
The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.
E_total = E + E₁
the expression for the field of a point charge is
E₁ = k q₁ / r²
for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative
E_total = E -k q₁ / r₂
we substitute
0 = E - k q₁ / r²
q₁ =
let's calculate
q₁ =
q₁ = 1.78 10⁻⁹ C
now we can calculate the field for position x = 4 m
E_total = 4 - 9 10⁹ 1.78 10⁻⁹ / 4²2
E_total = 3 N / A
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