...refrigerate temperature. 5.The foot of the cupboard uses the stainless steel to be in charge of , enclose level and adjust the foot. 6.Plywood and foot shelf are ...
... unit is Ampere。An ampere is one coulomb of charge going past one point in the duration of one second。 C. Voltage (electrical potential difference...
(a) 圓環取一小段視為點電荷,其電量為ρdl 在套用點電荷的電場公式(q/4πεR^2) dE= ρdl /4πεR^2 在(0,0,h)的電場平行於xy平面的部份會被消掉, 只剩下z方向的部份 dEz=(ρdl /4πεR^2) sinθ θ為 r與圓環半徑b的夾角...
...potential energy--- 位能 (電位差) charge -field system--- 電場particle--- 粒子因此,The charge in potential energy of the charge -field system when the test particle is moved between the ...
...1.80 mm thick, with a dielectric constant of k=3.60. The resultant electric field in the dielectric is E = 1.20×10^6 volts/m 1.Compute the magnitude of the charge per unit area σ on the conducting plate.σ = E/4πk= 1.2*10^6/4π...
A帶正電,移走電子後正電量增加將電子一道B上,負電量增加設移動的電量為q原來的吸力為F = k*Q1*Q2/r^2= 9*10^9 *(1*10^-6) (1*10^-6) / (0.02^2) = 22.5 (N)移動電子後F’ = 9*10^9 *(1*10^-6 + q) (1*10^-6 + q) / (0.02^2) = 45 (N)兩式相除(1*10^-6)^2 / (1*10^-6 + q...
1.電磁力F=電荷q*速度V*磁場B向心力F = mV^2/R平衡後,qVB = mV^2/RV = qBR/m動能E = 1/2*mV^2= 0.5 * m * (qBR/m)^2= (qBR)^2 / (2m)= (2* 1.6*10^-19 * 0.5 * 0.5)^2 /(2* 6.64 * 10^-27)= 4.82*10^-13 (J)= 4.82*10...