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Morgarella [4.7K]
3 years ago
9

Numerical

Physics
1 answer:
ira [324]3 years ago
5 0

Answer:

Thus, 4 kg mass must be suspended from a steel wire 2 m long and 1 mm diameter to stretch it by 1 mm.

Explanation:

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The charge density of radius 12.0 cm isgiven by with a = 1.40 μC/m 3 and r ismeasured radially outward from theorigin. What is t
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Answer:

The electric potential of the uniformly charge disk is 1392.1 V

Explanation:

Electric potential, for a uniformly charged disk at a distance A, is given as;

V = \frac{\sigma}{2 \epsilon} [\sqrt{A^2 +R^2} -A]

Where;

σ is the charge density = 1.40 μC/m³

ε is the permittivity of free space = 8.85 x 10⁻¹²

A is the distance above the disk = 40 cm = 0.4 m

R is the radius of the disk = 0.12 m

Substitute in these values into the equation above, we will have

V = \frac{1.4 X 10^{-6}}{2X8,85X10^{-12}}[\sqrt{0.4^2 +0.12^2}-0.4] \\\\V = (79096.05)(0.0176) = 1392.1 V

Therefore, the electric potential of the uniformly charge disk is 1392.1 V

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