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ohaa [14]
3 years ago
14

A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte

ry disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated?
Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

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1115560000 J

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Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and fricti
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15. Calculate The coefficient of kinetic friction be-
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Wt = 26.84 [N]

Explanation:

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In this case, we must sum the works of the force applied by the box and the friction force that also acts on the box.

The friction force is defined as the product of the normal force by the coefficient of friction.

f = N*μ

where:

N = normal force = m*g [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

f = friction force [N]

μ = friction coefficient = 0.21

f = 72*9.81*0.21

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Now the total work:

Wt = WF - Wf

where:

Wt = total work [J] (units of Joules)

WF = work by the pushing force [J]

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Note: The friction force exerts a negative work, because this force is acting in opposite direction to the movement, therefore the negative sign.

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Answer:

Q = 116.8 J

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3 0
3 years ago
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