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Yakvenalex [24]
3 years ago
14

The spacing between two closely spaced oppositely charged parallel plates is decreased. What happens to the electrostatic potent

ial difference between the plates, assuming they form an isolated system: (a) it increases, (b) it decreases, (c) it stays the same, or (d) you can’t tell from the information given?
Physics
1 answer:
erastovalidia [21]3 years ago
5 0
<h2>Electrostatic Potential Decreases</h2>

Explanation:

  • If the spacing between two closely spaced oppositely charged parallel plates is decreased the electrostatic potential difference between the plates will decrease.
  • An electrostatic potential that is also referred to as the electric field potential or potential drop is the amount of work required to replace a unit of charge from a reference point to a specific point inside the electric field without any change in acceleration.  
  • Therefore, if the distance will decrease between oppositely charged plates there will be more affinity to attract which will reduce the amount of work done thus decreasing the electric potential

∴  The Correct option is (b)

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Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.
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Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

     E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

   E1 = k (-q) / a²

   E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

    E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

    Et = E1 + E2

    Et = kq / a² + kq / a²

    Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

     E1 = k q / (R + a)²

     E2 = kq / (R-a)²

     Et = kq [1 / (R + a)² + 1 / (R-a)²]

     Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

     Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that  R> a we can despise in the patents "a"

     (R² + a²) = R² (1+ a² / R²) ≈ R²

     (R + a)² = R² (1 + a / R)² ≈ R²

     (R- a)²  = R² (1-a / R)² ≈ R²

Substituting in the total electric field

     Et = kq {2 R²) / [R²R²]}

     Et =kq 2 / R²

7 0
3 years ago
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Answer:

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8 0
2 years ago
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

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let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

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Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

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                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

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Therefore by balancing the forces, we get

Spring force= weight

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Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

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3 years ago
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