Struggling With Part C of Electric Field Calculation

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The discussion centers on a user's difficulty with Part C of an electric field calculation, specifically using the equation V = kQ1/r1 + kQ2/r2. The user believes their calculated answer of 8.329 V is incorrect but is unsure why. Other participants suggest that the user clarify their calculations and consider whether they are using the correct equation. They also recommend creating a sketch to visualize the relationship between electric field strength and distance. Overall, the thread emphasizes the importance of sharing detailed work for accurate troubleshooting.
johnio09
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Homework Statement
Consider a solid conducting sphere with a radius 1.5 cm and charge -4.4pC on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius 3.7 cm and outer radius 5.1 cm and a net charge 27.4 pC on the shell. A) denote the charge on the inner surface of the shell by Q'2 and that on the outer surface of the shell by Q ''2 . Find the charge Q''2. Answer in units of pC. B) Find the magnitude of the electric field at point P, midway between the outer surface of the solid conducting sphere and the inner surface of the conducting spherical shell. Answer in units N/C. C) Find the potential V at point P. Assume the potential at r = infinity. Answer in units of volt.
Relevant Equations
E =kQ/r^2
V = kQ/r
I've figured out parts A and B but I'm struggling with Part C. I used the equation V = kQ1/r1 + kQ2/r2 where Q1 = -4.4e-12C ; k = 8.98755e9 r1 = 0.026 m Q2 = 27.4e-12 and r2 = .051-.026 My answer (8.329 V) is wrong but I have no idea why. Please help if you can.
 
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johnio09 said:
My answer (8.329 V) is wrong but I have no idea why.
I have no idea either why your answer 8.329 V is wrong because my mind-reading abilities are not what they used to be. I can't help you find what's wrong unless you post what you did and how you got that answer.
 
Hello @johnio09 ,
:welcome: ##\qquad## !​

johnio09 said:
I've figured out parts A and B
Perhaps you can enlighten us ?

johnio09 said:
wrong but I have no idea why
How do you know it's wrong ? Because the book answer is different ?

Perhaps you used the wrong equation ? What's the idea behind it ?

Several ways out are feasible. My advice: make a sketch of ##|E| ## vs ##r##.

Oh, and do read the guidelines . Follow them as best you can and we'll get along just fine :smile: !

##\ ##
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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