I@Gyro Gearloose
INote for OP
+solved @user1 @user2...
ODirect Laplace equation
Sthe what eqn
O$$\nabla^{2}V = \frac{-\rho}{\epsilon}$$
OIn your case, since V doesn't depend on theta and phi, it simplifies
Sin which chap is this eqn in book i hve never seen this
OIt's Gauss' Law in differential form combined with E = -dV/dr
OOOPS
OWait
TOpt
Snever seen this i need to read up more on this stuff
OSorry, I missed the minus sign
OIt's fine, you can do without that
Sthnk god because partial differntiations just... scare me
OYou just do it in two steps basically
OHere because there's no theta or phi, it's a simple derivative only
Sthts great though
WThe hell is all this
WLaplace was in Thermodynamics
Si suck at simple diffrential at times
WIirc
O$$\frac{d^{2}V}{dr^{2}}=\frac{-\rho}{\epsilon}$$
TOpt
Sexct my rxn
Sso we integrate both sides to get the result here?
ONo, no need
OIt's rho that is asked
OSo differentiate V twice and multiply by epsilon
OShould give answer
Saah easy
Si will try now
WOpt bhai samaanya aam simple bhasha me bata do kuch
OOk, I'll explain it in a way that's mathematically wrong, but simple to understand
OE × A = q/e (I'll use e for epsilon)
Sif i am not diffrential dumb it should be 4a (epslon not)
WNo ye to gauss law hai
S
final ans bolri huOE × A × e = q
Divide by volume
E × e/l = rho
But E = -V/l
-V/l × e/l = rho
Rho = -e(d²V/dr²)
Dimensions are correct
Divide by volume
E × e/l = rho
But E = -V/l
-V/l × e/l = rho
Rho = -e(d²V/dr²)
Dimensions are correct
WWoah
WCrwazy
Smujhe isse samjhne do
S10 min lagega
WHow are supposed to figure this out in the exam again?
W
Sgood question
OWI fortunately haven't so yea lucky day
Si didnt get the last step
OActual explanation is that Gauss' Law, which is integral as taught to you guys, can also be written in a differential form.
