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2 years ago

7

What must be the distance in meters between point charge q1 = 30.3 µC and point charge q2 = -68.2 µC for the electrostatic force

between them to have a magnitude of 7.09 N?

1 answer:

Zolol [24]2 years ago

4 0

Answer: 1.62 m

Explanation: In order to solve this problem we have to use the definition of Coulomb force whic is given by:

F=(k*q1*q2)/d^2 where k is a constant 9*10^9 N/C^2*m^2. d is the distance between tha charges q1 and q2.

then we can:

d^2= (k*q1*q2)/F= (9*10^9*30.3*10^-6*68.2*10^-6)/7.09=1.62 m

(note we have considered q2 as positive because we are determining d^2 by using the magnitude of the electric Force bewteen the charges).

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Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

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A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0

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Answer:

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Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

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$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

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Hence the net torque produced by the two forces is given as a summation of both forces:

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b.) The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

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