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Yakvenalex [24]
2 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]2 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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The label on a battery-powered radio recommends the use of a rechargeable nickel-cadmium cell (nicads), although it has a 1.25-V
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Answer:

0.2 W more power than nicad cell is delivered by alkaline cell

Explanation:

<u>FOR NICKEL-CADMIUM CELL (nicads):</u>

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.25 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.04 Ω

Therefore,

I = (1.25 V)/(3.65 Ω + 0.04 Ω)

I = 0.34 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.34 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.34 A)²(3.65 Ω)

P = 0.41 W

<u>FOR ALKALINE CELL:</u>

First we find the current supplied to radio by the cell. For this purpose, we use the formula:

I = E/(R+r)

where,

I = current supplied

E = emf of cell = 1.58 V

R = resistance of radio = 3.65 Ω

r = internal resistance of cell = 0.2 Ω

Therefore,

I = (1.58 V)/(3.65 Ω + 0.2 Ω)

I = 0.41 A

Now, we calculate the power delivered to radio by following formula:

P = VI

but, from Ohm's Law:   V = IR

Therefore,

P = I²R

where,

P = Power delivered = ?

I = current = 0.41 A

R = Resistance of radio = 3.65 Ω

Therefore,

P = (0.41 A)²(3.65 Ω)

P = 0.61 W

Now, fo the difference between delivered powers by both cells:

ΔP = (P)alkaline - (P)nicad

ΔP = 0.61 W - 0.41 W

<u>ΔP = 0.2 W</u>

4 0
3 years ago
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