Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 7.00 % C. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be V = 0 at r = 0

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aristocles aristocles · Answer 1 · 2019-10-04T02:51:34+02:00

Answer:

Part a)

[tex]Q_1 = 2\mu C[/tex]

Part b)

[tex]\Delta V = 4.5 \times 10^4 V[/tex]

Explanation:

As we know that conducting sphere is treated like spherical capacitor

so we can say

[tex]\frac{C_1}{C_2} = \frac{0.4}{1}[/tex]

also we know that

[tex]Q = CV[/tex]

so charge on two spheres will be in the ratio of their capacitance

Part a)

[tex]Q_1 + Q_2 = 7 \mu C[/tex]

[tex]\frac{Q_1}{Q_2}= 0.4[/tex]

[tex]1.4Q_2 = 7\mu C[/tex]

[tex]Q_2 = 5 \mu C[/tex]

[tex]Q_1 = 2\mu C[/tex]

Part b)

Now potential difference between two sphere can be given as

[tex]\Delta V = \frac{Q}{C}[/tex]

[tex]\Delta V = \frac{2\mu C}{4\pi \epsilon_0 R_1}[/tex]

[tex]\Delta V = \frac{2\mu C}{4\pi \epsilon_0(0.400)}[/tex]

[tex]\Delta V = 4.5 \times 10^4 V[/tex]