Yo tengo quince y soy de cali y estoy viviendo en mexico
Answer:
The answer is not able to be solved, because we dont know what objects are in it, and how heavy they are. More information please!
Explanation:
Answer:
a)![E=50.53\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D50.53%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be negative direction.
b)
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
![E=K\dfrac{q}{r^2}](https://tex.z-dn.net/?f=E%3DK%5Cdfrac%7Bq%7D%7Br%5E2%7D)
![E_1=K\dfrac{q_1}{r_1^2}](https://tex.z-dn.net/?f=E_1%3DK%5Cdfrac%7Bq_1%7D%7Br_1%5E2%7D)
![E_2=K\dfrac{q_2}{r_2^2}](https://tex.z-dn.net/?f=E_2%3DK%5Cdfrac%7Bq_2%7D%7Br_2%5E2%7D)
Now by putting the va;ues
a)
![E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B7.7%5Ctimes%2010%5E%7B-6%7D%7D%7B0.031%5E2%7D%5C%20N%2FC)
![E_1=72.11\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_1%3D72.11%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
![E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C](https://tex.z-dn.net/?f=E_2%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B19%5Ctimes%2010%5E%7B-6%7D%7D%7B0.089%5E2%7D%5C%20N%2FC)
![E_2=21.58\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_2%3D21.58%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The net electric field
![E=E_1-E_2](https://tex.z-dn.net/?f=E%3DE_1-E_2)
![E=50.53\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D50.53%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
![E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C](https://tex.z-dn.net/?f=E_1%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B7.7%5Ctimes%2010%5E%7B-6%7D%7D%7B0.024%5E2%7D%5C%20N%2FC)
![E_1=120.3\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_1%3D120.3%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
![E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C](https://tex.z-dn.net/?f=E_2%3D9%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B19%5Ctimes%2010%5E%7B-6%7D%7D%7B0.034%5E2%7D%5C%20N%2FC)
![E_2=147.92\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E_2%3D147.92%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The net electric field
![E=E_1+E_2](https://tex.z-dn.net/?f=E%3DE_1%2BE_2)
![E=268.22\times 10^{6}\ N/C](https://tex.z-dn.net/?f=E%3D268.22%5Ctimes%2010%5E%7B6%7D%5C%20N%2FC)
The direction will be positive direction.
Answer:
i think the answer is 12 ohms
plz mark me as brainliest :)