Re: [題目] 最近剛開始準備普物轉學考 有六題靜電學的不會 請幫幫我
※ 引述《janelin319 (小人)》之銘言：
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讓我試試看這樣解釋看能不能有頭緒一點
第一題我想他的a指的應該是半徑,然後她的axis指的應該是穿過圓的中心點
the answer following will be using cgs unit instead of MKS(SI) unit
the expression will be a little difference but its the same thing
Im trying to use English to explain just beacause there are some
terminology i dont know how to say it in chinese,but I will try my best
→
therefore, dE=(rho/r^3) r d^3r (rho=charge density=Q/2 a pi)
(r=(x^2+a^2)^1/2)
(d^3r=adtheta)
E=Integral_______rho____(x,0,0)ad(theta)
(x^2+a^2)^3/2
Integrate from [0,2pi]
after integration, you will find a general solution for E-field
next, take the derivative to maximize (with respect to x)
set the derivative=0, you will find x and sub it back to the
expression you will find value of Emax
第二題
assuming x is along vertical axis and y is along horizontal axis
use the same expreesion to start it from the first part
→
dE=(rho/r^3) r d^3r
(rho=Q/L) Q is the charge L is the length of the rod
(r=L/pi)r is the radius of the semicircle
→
(r =(-cos(theta)L/pi),-sin(theta)L/pi)
and then integrate it from[0,pi]
第三題
again using the same expression to start with
btw, the hardest part just set up the integral and the rest are pretty
easy
charge density=lambda (we use rho before just to generalize it in 3D)
r=(x^2+y^2)^1/2
→
r=(0,y)
dx=dx
after you set up the integral integrate it from[-L/2,L/2]
at some point you will use trig substitution that is given in the question
the answer you will find contain expression of sine (ie sin(theta1)-sin(the2))
, and by taking the limit you will find it becomes a fator of 2, which gives
you the expression for infinite line
第四題
this one by the given hint,
first, break the cylinder into 3 parts, side+2 cover
use the ring as the "side" of the cylinder (from question 1 we have the
expression of electric field of the ring) and the length of the cylinder
is L. So, the density contains in each dL is L/n (n is the number of
density of the ring contain in each dL)
Next, using the same procedure for dE, find the E for 2 cover, which is just
two circular plate. I will give you some parameter here
rho=2Rpi(R+L)
r=(r'^2+z^2)^1/2
→
r=(0,0,z) (we are assuming its along z axis)
d^3r=r'dr'dtheta
and you just have to integrate over r=[0,R] theta=[0,2pi]
第五題
this is the question of Fg=Fq
mgsin(theta)=qE
E=mg sin(theta)/q
and you can figure out the direction from the diagram
第六題
(a) its really hard to draw lines here so ill pass the first part and
the second part find the place E=0
From the symmetry, you can tell it is zero at the center but to be more
qualitative, draw x and y axis through the center of triangle
break E field for each charge into component and sum it up
(b) use the axis you drew from part a), you can figure out the component
and just use the superposition principle Etotal=E1+E2
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03/13 21:21, , 1F
03/13 21:21, 1F