Calculate Charge: Find Q at Center of Hollow Sphere of Radius 20 cm

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The discussion centers on calculating the unknown charge Q located at the center of a hollow conducting sphere with a radius of 20 cm. The electric field just outside the sphere is 500 N/C, while the field at a point further out is 100 N/C. By applying Gauss's law, it is determined that the total charge on the conducting shell is 2.225 nC, and the presence of charge Q alters the net charge within the Gaussian surface, leading to the conclusion about its magnitude and sign.

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austin007
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The e-field just outside the surface of a hollow conducting sphere of radius 20 cm is 500 N/C and is directed outward. Q is located at the center of the sphere. It is noted that the e-fields is still directed out ward with a magnitude of 100 N/C. What is the magnitude and sign of the unknown charge Q?

I can only find total charge which is 2.225nC. I can't get Q coz radius of the solide sphere is unknown. Pls help
 
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Can you please state the problem completely, exactly as it was given. I don't understand what the field 100 N/C refers to.
 
The electric field just outside the surface of a hollow conducting sphere of radius 20 cm is 500 N/C and is directed outward. An unknown charge Q is then located at the center of the sphere. It is noted that the electric field is still directed outward but its magnitude has decreased to 100 N/C. What is the magnitude and sign of the unknown charge Q?

This is the complete question as it is stated.
 
austin007 said:
The electric field just outside the surface of a hollow conducting sphere of radius 20 cm is 500 N/C and is directed outward.
OK. That should tell you the charge on the conducting shell.
An unknown charge Q is then located at the center of the sphere.
Ah... that additional word "then" makes all the difference!
It is noted that the electric field is still directed outward with a magnitude of 100 N/C.
Now that you've added charge Q, how does the net charge within the gaussian surface (taken just outside the sphere) change?

Apply Gauss's law twice.
 
Last edited:
thanks I understood finally...
 

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