A 2.0 -cm-radius metal sphere carries \(75 \mathrm{nC}\) and is surrounded by a concentric spherical conducting shell of radius \(10 \mathrm{cm}\) carrying -75 nC. (a) Find the potential difference between shell and sphere. (b) How would your answer change if the shell's charge were +150 nC?

Short Answer

Expert verified
The initial potential difference is determined by the charges on the sphere and the shell. However, when the charge on the shell increases to \(+150 nC\), the potential difference is affected and may increase or decrease depending on whether the charge on the spherical shell is less or more than the charge on the metal sphere.

Step by step solution

01

Finding initial potential difference

First, the potential \(V_s\) of the small sphere is given by \(V_s = k*q_s/r_s\) where \(k = 9.0*10^9 Nm^2/C^2\) is the Coulomb's constant, \(q_s = 75 nC\) is the charge of the sphere and \(r_s = 2 cm\) is the radius of the sphere. Secondly, the potential \(V_b\) of the shell is given by \(V_b = k*q_b/r_b\) where \(q_b = -75 nC\) is the charge of the conducting shell and \(r_b = 10 cm\) is the radius of the shell. Then, calculate the potential difference \(V_{sb} = |V_s - V_b|\).
02

Finding new potential difference with different shell charge

Now, the problem is to find the potential difference if the charge of the shell were \(+150 nC\). We calculate the new potential on the shell \(V'_b = k*q'_b/r_b\) where \(q'_b = 150 nC\). After calculating \(V'_b\), find the new potential difference \(V'_{sb} = |V_s - V'_b|\).
03

Comparing the potential differences

We can compare the initial potential difference \(V_{sb}\) with the new potential difference \(V'_{sb}\) to see how it changes when the shell's charge is \(+150 nC\).

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