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Ksivusya [100]
4 years ago
13

An isolated conducting sphere has a 16 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

ent of 1.0000000 A out of it. How long would it take for the sphere to increase in potential by 1260 V?
Physics
1 answer:
timofeeve [1]4 years ago
5 0

Answer:

It takes 11.20 ms for the sphere to increase in potential by 1260 V

Explanation:

Using the formula q = It where I is current in Ampere, and t is in seconds

q = ( 1.000002 - 1.0000000) t

q = 0.000002t

Voltage on the surface of the sphere V = Kq / r

where K is 9.0 × 10⁹ N².m² / c² and

R = 16 cm = 16 /100 = 0.16m

V = Kq /r

substitute the value into q

V = K(0.000002t) / r

cross multiply

rV = K × 0.000002t

make t subject of the formula

t = 0.16 × 1260 / ( 9×10⁹ × 0.000002)

t = 201.6 / ( 18 × 10³)

t = 0.0112 s

  = 11.20 ms

It takes 11.20 ms for the sphere to increase in potential by 1260 V

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How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele
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Explanation:

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V₂ = Electric Potential at Final Position = - 8.9 V

Therefore,

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Since, we use magnitude in calculation only. Therefore,

ΔV = 15.6 V

Now, we calculate total charge:

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