The field just outside a metal ball

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The electric field just outside a 3.22 cm radius metal ball is measured at 2.72×10^2 N/C, directed toward the ball. Using the equation E = k*q/r^2, the calculated charge on the ball is approximately 3.14×10^-11 C. However, the field's direction indicates that the charge is negative, as it points toward the ball. This aligns with Gauss's law, which supports the conclusion of a negative charge. Therefore, the correct interpretation of the charge is essential for accurate problem-solving.
Hooke's Law
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Homework Statement



The field just outside a 3.22 cm radius metal ball is 2.72×10^2N/C and points toward the ball. What charge resides on the ball?

Homework Equations



E = k*q/r^2



The Attempt at a Solution



q = E*r^2/k

q = ( 2.72*10^2 N/C )* (0.0322 m^2) / 8.99*10^9

= 3.14*10^-11 C


The computer says I'm giving it the wrong answer...is there something wrong with what I did?


Thanks
 
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The magnitude of your answer seems correct...thats what Gauss's law gives ...
but i think the charge is negative because the field is pointing towards the ball...try it...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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