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Levart [38]
3 years ago
8

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

nd positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.
Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?
Physics
1 answer:
Yanka [14]3 years ago
4 0

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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Answer:

0.9 meters

Solution:

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First one is Force Balance and second one is Torque Balance.

<u>Force Balance</u>

First equation is generally used to know the magnitude of the force applied.

Since they have to lift wooden board in order to carry it, therefore they have to apply a counter force to overcome its weight, which is 160 N.

So the net force applied by the two men in upward direction is 160 N.

The force applied by person 1 (as in the figure attached) is 60 N and by person 2 is F N. So,

60 + F = 160

∴ F = 100 N

<u>Torque Balance</u>

(<em>Torque is nothing but a twisting force that tends to cause rotation.</em>

<em>To calculate torque multiply the force applied with the perpendicular distance between axis of rotation and line of application of force</em>)

The second equation is generally used to know the position of the force applied.

Since they are carrying wooden board in horizontal position, therefore there is no net torque on the wooden board.

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Since torque of two person are opposing each other as one is in clockwise direction and other one is in anticlockwise direction, therefore, if they are equal then the net torque will become zero. So,

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x = \frac{90}{100}

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