I think it’s B hope it helps:)
Answer:
2.2 meters
Explanation:
Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:
![KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7BkQq%7D%7Br%7D%5C%20%20%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20...i)
is Coulomb's constant.
#The electric field,
at radius r is expressed as:
![E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQ%7D%7Br%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20...ii)
From i and ii, we have:
![KE=Eqr](https://tex.z-dn.net/?f=KE%3DEqr)
![r=(KE)/Eq](https://tex.z-dn.net/?f=r%3D%28KE%29%2FEq)
#Substitute actual values in our equation:
![r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B75J%7D%7B%287.2%5Ctimes%2010%5E%7B-5%7DC%29%284.8%5Ctimes%2010%5E5%20V%2Fm%29%7D%5C%5C%5C%5C%3D2.1701%5Capprox2.2%5C%20m)
Hence, the distance between the charge and the source of the electric field is 2.2 meters
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so
![Q_{conduction} = Q_{convection}](https://tex.z-dn.net/?f=Q_%7Bconduction%7D%20%3D%20Q_%7Bconvection%7D)
Expression for the heat conduction process is
![Q_{conduction} = \frac{K(T_1 - T)}{L}](https://tex.z-dn.net/?f=Q_%7Bconduction%7D%20%3D%20%5Cfrac%7BK%28T_1%20-%20T%29%7D%7BL%7D)
Expression for the heat convection process is
![Q_{convection} = h(T_2 - T)](https://tex.z-dn.net/?f=Q_%7Bconvection%7D%20%3D%20h%28T_2%20-%20T%29)
Substitute the expressions of conduction and convection in equation above
![Q_{conduction} = Q_{convection}](https://tex.z-dn.net/?f=Q_%7Bconduction%7D%20%3D%20Q_%7Bconvection%7D)
![\frac{K(T_1 - T_2)}{L} = h(T_2 - T)](https://tex.z-dn.net/?f=%5Cfrac%7BK%28T_1%20-%20T_2%29%7D%7BL%7D%20%3D%20h%28T_2%20-%20T%29)
Substitute the values in above equation
![\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC](https://tex.z-dn.net/?f=%5Cfrac%7B2.79%2850-%20T_2%29%7D%7B0.2%7D%20%3D%2015%28T_2%20-%2022%29%5C%5C%5C%5CT_2%20%3D%2035.5%5E%5CcircC)
Now heat flux through the wall can be calculated as
![q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2](https://tex.z-dn.net/?f=q_%7Bflux%7D%20%3D%20Q_%7Bconduction%7D%20%5C%5C%5C%5Cq_%7Bflux%7D%20%20%3D%20%5Cfrac%7BK%28T_1%20-%20T_2%29%7D%7BL%7D%5C%5C%5C%5Cq_%7Bflux%7D%20%20%3D%20%5Cfrac%7B2.79%2850%20-%2035.5%29%7D%7B0.2%7D%5C%5C%5C%5Cq_%7Bflux%7D%20%3D%20202.3W%2Fm%5E2)
Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²