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spin [16.1K]
3 years ago
12

Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric fi

eld E between the plates? Express E in terms of q and other quantities given in the introduction, in addition to ϵ0 and any other constants needed..
Physics
1 answer:
Anna [14]3 years ago
8 0

Answer:

 E=\frac{q}{\epsilon_oA}  

Explanation:

Let the area of each plate be A.

From Gauss Law,

\phi=\frac{q}{\epsilon_o}

\phi=E.A

\Rightarrow E.A=\frac{q}{\epsilon_o}\\\Rightarrow E=\frac{q}{\epsilon_oA}

Thus, using Gauss law, electric field between the plates is:

E=\frac{q}{\epsilon_oA}

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Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

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