Answer:
![E_{net} = 6.44 \times 10^5 N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%20%3D%206.44%20%5Ctimes%2010%5E5%20N%2FC)
Explanation:
As we know that electric field due to infinite line charge distribution at some distance from it is given as
![E = \frac{2k \lambda}{r}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B2k%20%5Clambda%7D%7Br%7D)
now we need to find the electric field at mid point of two wires
So here we need to add the field due to two wires as they are oppositely charged
Now we will have
![E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}](https://tex.z-dn.net/?f=E_%7Bnet%7D%20%3D%20%5Cfrac%7B2k%5Clambda_1%7D%7Br%7D%20%2B%20%5Cfrac%7B2k%5Clambda_2%7D%7Br%7D)
now plug in all data
![\lambda_1 = 4.68 \muC/m](https://tex.z-dn.net/?f=%5Clambda_1%20%3D%204.68%20%5CmuC%2Fm)
![\lambda_2 = 2.48 \mu C/m](https://tex.z-dn.net/?f=%5Clambda_2%20%3D%202.48%20%5Cmu%20C%2Fm)
![r = 0.200 m](https://tex.z-dn.net/?f=r%20%3D%200.200%20m)
now we have
![E_{net} = \frac{2k}{r}(4.68 + 2.48)](https://tex.z-dn.net/?f=E_%7Bnet%7D%20%3D%20%5Cfrac%7B2k%7D%7Br%7D%284.68%20%2B%202.48%29)
![E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})](https://tex.z-dn.net/?f=E_%7Bnet%7D%20%3D%20%5Cfrac%7B2%289%5Ctimes%2010%5E9%29%7D%7B0.200%7D%287.16%20%5Ctimes%2010%5E%7B-6%7D%29)
![E_{net} = 6.44 \times 10^5 N/C](https://tex.z-dn.net/?f=E_%7Bnet%7D%20%3D%206.44%20%5Ctimes%2010%5E5%20N%2FC)
Pet rocks contain organic matter
Answer:
sorry i throght i had the answer
Explanation:
Answer:
The amount of work done required to stretch spring by additional 4 cm is 64 J.
Explanation:
The energy used for stretching spring is given by the relation :
.......(1)
Here k is spring constant and x is the displacement of spring from its equilibrium position.
For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).
![8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}](https://tex.z-dn.net/?f=8%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimesk%5Ctimes%20k%20%5Ctimes%280.02%29%5E%7B2%7D)
k = 4 x 10⁴ N/m
Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).
Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).
![E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes4%5Ctimes10%5E%7B4%7D%5Ctimes%20%280.06%29%5E%7B2%7D)
E = 72 J
But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :
E = ( 72 - 8 ) J = 64 J