You are attracting electricity<span />
Answer:
Part a)
![W = 1.58 \times 10^{-20} J](https://tex.z-dn.net/?f=W%20%3D%201.58%20%5Ctimes%2010%5E%7B-20%7D%20J)
Part b)
![v = 1.86 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%201.86%20%5Ctimes%2010%5E5%20m%2Fs)
Explanation:
Part a)
Electric field due to large sheet is given as
![E = \frac{\sigma}{2\epsilon_0}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5Csigma%7D%7B2%5Cepsilon_0%7D)
![\sigma = 4.00 \times 10^{-12} C/m^2](https://tex.z-dn.net/?f=%5Csigma%20%3D%204.00%20%5Ctimes%2010%5E%7B-12%7D%20C%2Fm%5E2)
now the electric field is given as
![E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B4.00%20%5Ctimes%2010%5E%7B-12%7D%7D%7B2%288.85%20%5Ctimes%2010%5E%7B-12%7D%29%7D)
![E = 0.225 N/C](https://tex.z-dn.net/?f=E%20%3D%200.225%20N%2FC)
Now acceleration of an electron due to this electric field is given as
![a = \frac{eE}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BeE%7D%7Bm%7D)
![a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%280.225%29%7D%7B9.1%20%5Ctimes%2010%5E%7B-31%7D%7D)
![a = 3.97 \times 10^{10}](https://tex.z-dn.net/?f=a%20%3D%203.97%20%5Ctimes%2010%5E%7B10%7D)
Now work done on the electron due to this electric field
![W = F.d](https://tex.z-dn.net/?f=W%20%3D%20F.d)
![d = 0.470 - 0.03](https://tex.z-dn.net/?f=d%20%3D%200.470%20-%200.03)
![d = 0.44 m](https://tex.z-dn.net/?f=d%20%3D%200.44%20m)
So work done is given as
![W = (ma)(0.44)](https://tex.z-dn.net/?f=W%20%3D%20%28ma%29%280.44%29)
![W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)](https://tex.z-dn.net/?f=W%20%3D%20%289.11%20%5Ctimes%2010%5E%7B-31%7D%29%283.97%20%5Ctimes%2010%5E%7B10%7D%29%280.44%29)
![W = 1.58 \times 10^{-20} J](https://tex.z-dn.net/?f=W%20%3D%201.58%20%5Ctimes%2010%5E%7B-20%7D%20J)
Part b)
Now we know that work done by all forces = change in kinetic energy of the electron
so we will have
![W = \frac{1}{2}mv^2 - 0](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%200)
![1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2](https://tex.z-dn.net/?f=1.58%20%5Ctimes%2010%5E%7B-20%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%289.1%5Ctimes%2010%5E%7B-31%7D%29v%5E2)
![v = 1.86 \times 10^5 m/s](https://tex.z-dn.net/?f=v%20%3D%201.86%20%5Ctimes%2010%5E5%20m%2Fs)
<span>a decrease in the distance between the earth and the moon</span>
Answer:
d = (3 i + 5 j) 3 [m] x-distance and 5 [m] y-distance
Explanation:
We can find the displacement if we use a diagram with the cartesian coordinates of the displacement of the mouse.
In the attached image we can see each of the vectors and the origin where the mouse was in the initial position and then the final position after the mouse moved to the right.
So the mouse moved 3 meters to the right and 5 meter upward.