How Is Charge Determined from Electric Potential and Distance?

AI Thread Summary
The electric potential of 33.0 V at a distance of 20.5 cm from a charged sphere allows for the calculation of the sphere's charge using the formula q = V * r / k, where k is the Coulomb's constant. There was a clarification that 20.5 cm should be correctly converted to meters for accurate calculations, as 0.255 m is incorrect. The discussion emphasizes that the potential does not need conversion into different units since it is already in volts. Participants confirm the approach to solving the problem is valid and highlight the importance of unit accuracy. The thread concludes with a focus on correctly applying the formula to find the charge.
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Homework Statement


The electric potential at a distance of 20.5 cm from a very small charged sphere is 33.0 V, with the potential taken to be zero at an infinite distance from the sphere.

If the sphere is treated as a point charge, what is its charge?

Homework Equations



V = \frac {U} {q_{0}} == \frac {1} {4*pi*(8.85*10^{-12})}*\frac {q} {r}

The Attempt at a Solution



How do you convert 33.0V to the correct units of Q or is it equliv?

\frac {1} {4*pi*(8.85*10^{-12})}*\frac {33.0} {0.255} ??
 
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You don't need to convert your potential into anything, you already have an expression for V, so use it;

V = k\cdot\frac{q}{r} \Leftrightarrow q = \frac{V\cdot r}{k}

Also note that 20.5 cm \neq 0.255 m, its probably just a typo.
 
Hootenanny said:
You don't need to convert your potential into anything, you already have an expression for V, so use it;

V = k\cdot\frac{q}{r} \Leftrightarrow q = \frac{V\cdot r}{k}

Also note that 20.5 cm \neq 0.255 m, its probably just a typo.

Thank you, yes that was a typo.
 
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