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Paha777 [63]
3 years ago
12

To give an idea of sensitivity of the platypus's electric sense, how far from a 80nC n C point charge does the field have this m

agnitude
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

The question is incomplete. The complete question is :

A platypus foraging for prey can detect an electric field as small as 0.002 N/C.

-To give an idea of sensitivity of the platypus's electric sense, how far from a +80nC point charge does the field have this magnitude?

Solution :

Given electric field,  E = 0.002 N/C

Charge, Q = + 80 nC

$\therefore E = \frac{kQ}{R^2} $

or $R^2=\frac{kQ}{E}$

    $R^2=\frac{9\times 10^9 \times 80 \times 10^{-9}}{0.002}$

   R = 600 m

This is the distance of the charge from the point of observations.

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In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

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F = qE

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q is charge in C

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Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

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mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

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F = qE

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Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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