How Long Does It Take for a Copper Sphere's Potential to Increase by 1000V?

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Homework Help Overview

The discussion revolves around the time required for a copper sphere's electric potential to increase by 1000V, considering its radioactive nickel coating that emits electrons. The problem involves concepts from electrostatics and radioactivity, particularly how charge accumulation affects electric potential.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between electric potential and charge, with one questioning the definition of potential in this context. Others suggest using electric field equations to relate potential and charge, while also considering the uniformity of potential across the sphere's surface.

Discussion Status

Some participants have provided insights into the calculations needed to determine the charge required for the potential increase. There is an ongoing examination of the formulas and assumptions related to the electric field and potential of a charged conducting sphere, with no explicit consensus reached yet.

Contextual Notes

Participants note potential errors in the application of formulas related to the electric field outside a charged conducting sphere, indicating a need for careful consideration of the correct equations and assumptions in the problem setup.

Hyperreality
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A solid copper sphere whose radius is 1.0cm has a very thin surface coating of nickel. Some of the nickel atoms are radioactive and emit an electron with each decay. Due to the geometry of the situation, half of these electrons enter the copper sphere, each carrying away a charge of -e. The nickel coating has an activity of 10mCi(millicuries) = 3.70 x 10^8 radioactive decays per second. The sphere is hung from a long, nonconducting string and isolated from its surroundings.

How long will it takefor the potential of the sphere to increase by 1000V?

While it is easy to work out the net charge of copper = the no. of -e decayed, and the energy gained by copper, I don't really know what they meant by potentials of the sphere in this case. Since potential has unit of Joules / Coulomb, does it mean that in this case,

Potentials = Total Energy Gain / Total charge of the sphere??
 
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Potential is the energy required (to charge)per unit charge or dU/dq
 
I had a think of the problem, afterwards. Below is my working. Potentials is created via the existence of an electric field. So, the electric field needed is

E = V / r.
V = 1000, r = 0.01m
Then put

E = kQ / r and obtain a value for the total net charge Q required.

Then use the Q / e to find the number of electrons.

Is this the correct way in solving the problem? Here I'm assuming that the surface potential are same at all point.
 
Hyperreality said:
I had a think of the problem, afterwards. Below is my working. Potentials is created via the existence of an electric field. So, the electric field needed is

E = V / r.
V = 1000, r = 0.01m
Then put

E = kQ / r and obtain a value for the total net charge Q required.

Then use the Q / e to find the number of electrons.

Is this the correct way in solving the problem? Here I'm assuming that the surface potential are same at all point.
there's an error in your formula for E outside a charged conducting sphere.
it should be:
E = kQ/r2
see the URL links below for the potential of a charged conducting sphere and how to calculate it from the E field. (then you can determine Q and Q/e like you indicated above.)
http://230nsc1.phy-astr.gsu.edu/hbase/electric/potsph.html#c1
http://230nsc1.phy-astr.gsu.edu/hbase/electric/potpoi.html#c2
 
Last edited:

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