- #1
phenolic
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1. Point charge in the presence of a grounded conducting sphere given the boundary conditions phi(x)=0 at r=a (V=0) and at r=infinity (V=0)
2. The equation I used phi(x)= (q/abs(x-y))+(q'/abs(x-y))=(q/abs(x*nhatsubx-y'nhatsuby))+ (q'/abs(x'*nhatsubx-y'*nhatsuby')
3. The equation listed above is what I came up with at r=a q/abs(anhat-yn'hat)+q'/abs(a*nhat-y'*n'hat)=0, my next equation was q/a*abs(nhatsubx-(y/a)nhatsuby')+q'/y'abs(n'hatsuby-(a/y')nhatsubx). After solving they come up with the ratio (q/a)=-(q'/y') and (y/a)=(a/y'). Is it because of the symmetry that they are using these ratios as opposed to going through the rigormorale of law of cosines?
2. The equation I used phi(x)= (q/abs(x-y))+(q'/abs(x-y))=(q/abs(x*nhatsubx-y'nhatsuby))+ (q'/abs(x'*nhatsubx-y'*nhatsuby')
3. The equation listed above is what I came up with at r=a q/abs(anhat-yn'hat)+q'/abs(a*nhat-y'*n'hat)=0, my next equation was q/a*abs(nhatsubx-(y/a)nhatsuby')+q'/y'abs(n'hatsuby-(a/y')nhatsubx). After solving they come up with the ratio (q/a)=-(q'/y') and (y/a)=(a/y'). Is it because of the symmetry that they are using these ratios as opposed to going through the rigormorale of law of cosines?