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The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i =
where R is the resistance
R =
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i =
q =
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
Answer:
it will take for the sphere to increase in potential by 1500 V, 503.71 s.
Explanation:
The charge on the sphere after t seconds is:
q = (1.0000049 - 1.0000000) t = 0.0000049 t
The voltage on the surface is
V = k * = k 0.0000049 t / R
solve for t
t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * ) = 503.71 s