Potenziale guscio sferico
Salve a tutti,
mi sono iscritto appositamente per avere una risposta ad un quesito che mi è rimasto insoluto nel tempo (pur avendo dato l'esame di Fisica II all'uni) dato che il professore non ha mai chiarito in aula. Ricordo è stata anche una domanda oggetto di esame
Mi chiedeva di calcolare il potenziale di un guscio sferico (carica distribuita uniformemente sulla superficie) all'esterno della sfera ma esplicitando i calcoli.
Sappiamo che
$V(P) = Q/(4piepsilon_0x)$ (poichè $ E = - grad V $ e vabbè..)
cioè dipende dalla distanza $x$ del punto P dall'origine
Invece facciamo un altro ragionamento:
Posizioniamo il sistema di riferimento con l'asse z passante per il punto P e utilizziamo coordinate sferiche
[jxg]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[/jxg]
$dV = (dq)/(4piepsilon_0r) = (sigmadSigma)/(4piepsilon_0sqrt(R^2+x^2-2Rxcostheta) ) rArr V(P) = int (sigmadSigma)/(4piepsilon_0sqrt(R^2+x^2-2Rxcostheta)) = (sigmaR^2)/(4piepsilon_0) int _0^(2pi)int_0^pi (sinthetad\theta dphi)/sqrt(R^2+x^2-2Rxcostheta) = (sigmaR^2 2pi)/(4piepsilon_0) int_0^pi (sinthetad\theta)/(sqrt(2Rx)sqrt((R^2+x^2)/(2Rx)-costheta)) = (sigmaR^2)/(2epsilon_0sqrt(2Rx)) int_0^pi (sinthetad\theta)/sqrt(b-costheta) = $
( ponendo $ y = costheta rArr dy = -sinthetad\theta$ )
$ = -(sigmaR^2)/(2epsilon_0sqrt(2Rx)) int dy/sqrt(b-y) = -(sigmaR^2)/(2epsilon_0sqrt(2Rx)) [-2(b-y)^(1/2)] = (sigmaR^2)/(epsilon_0sqrt(2Rx)) [sqrt(b-costheta)]|_0^pi = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt(b+1) - sqrt(b-1)) = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt((R^2+x^2)/(2Rx)+1) - sqrt((R^2+x^2)/(2Rx)-1)) = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt((R+x)^2/(2Rx))-sqrt((R-x)^2/(2Rx))) = (sigmaR^2(R+x))/(epsilon_0\2Rx)-(sigmaR^2(R-x))/(epsilon_0\2Rx) = (2sigmaR^2x)/(2epsilon_0Rx) = (sigmaR)/epsilon_0 Sigma/Sigma = (QR)/(epsilon_0Sigma) = (QR)/(epsilon_o\4piR^2)$
$V(P) = Q/(4piepsilon_0R) $ con R raggio della sfera (costante) cioè non dipende dalla distanza del punto P dall'O.
Avete idea di dove sia l'errore?
mi sono iscritto appositamente per avere una risposta ad un quesito che mi è rimasto insoluto nel tempo (pur avendo dato l'esame di Fisica II all'uni) dato che il professore non ha mai chiarito in aula. Ricordo è stata anche una domanda oggetto di esame
Mi chiedeva di calcolare il potenziale di un guscio sferico (carica distribuita uniformemente sulla superficie) all'esterno della sfera ma esplicitando i calcoli.
Sappiamo che
$V(P) = Q/(4piepsilon_0x)$ (poichè $ E = - grad V $ e vabbè..)
cioè dipende dalla distanza $x$ del punto P dall'origine
Invece facciamo un altro ragionamento:
Posizioniamo il sistema di riferimento con l'asse z passante per il punto P e utilizziamo coordinate sferiche
[jxg]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[/jxg]
$dV = (dq)/(4piepsilon_0r) = (sigmadSigma)/(4piepsilon_0sqrt(R^2+x^2-2Rxcostheta) ) rArr V(P) = int (sigmadSigma)/(4piepsilon_0sqrt(R^2+x^2-2Rxcostheta)) = (sigmaR^2)/(4piepsilon_0) int _0^(2pi)int_0^pi (sinthetad\theta dphi)/sqrt(R^2+x^2-2Rxcostheta) = (sigmaR^2 2pi)/(4piepsilon_0) int_0^pi (sinthetad\theta)/(sqrt(2Rx)sqrt((R^2+x^2)/(2Rx)-costheta)) = (sigmaR^2)/(2epsilon_0sqrt(2Rx)) int_0^pi (sinthetad\theta)/sqrt(b-costheta) = $
( ponendo $ y = costheta rArr dy = -sinthetad\theta$ )
$ = -(sigmaR^2)/(2epsilon_0sqrt(2Rx)) int dy/sqrt(b-y) = -(sigmaR^2)/(2epsilon_0sqrt(2Rx)) [-2(b-y)^(1/2)] = (sigmaR^2)/(epsilon_0sqrt(2Rx)) [sqrt(b-costheta)]|_0^pi = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt(b+1) - sqrt(b-1)) = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt((R^2+x^2)/(2Rx)+1) - sqrt((R^2+x^2)/(2Rx)-1)) = (sigmaR^2)/(epsilon_0sqrt(2Rx))(sqrt((R+x)^2/(2Rx))-sqrt((R-x)^2/(2Rx))) = (sigmaR^2(R+x))/(epsilon_0\2Rx)-(sigmaR^2(R-x))/(epsilon_0\2Rx) = (2sigmaR^2x)/(2epsilon_0Rx) = (sigmaR)/epsilon_0 Sigma/Sigma = (QR)/(epsilon_0Sigma) = (QR)/(epsilon_o\4piR^2)$
$V(P) = Q/(4piepsilon_0R) $ con R raggio della sfera (costante) cioè non dipende dalla distanza del punto P dall'O.
Avete idea di dove sia l'errore?
Risposte
Proprio nessuno riesce a darmi una risposta? Se non la soluzione al problema, anche una considerazione?
E poi, come mai il professore ci presenta il problema per poi lasciarci nel dubbio?
E poi, come mai il professore ci presenta il problema per poi lasciarci nel dubbio?
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo