The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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Answer:
A) Cold object will start getting hot
B) Heat exchange will stop as the two object acquire the same temperature.
Explanation:
A) When one hot object and one cold object are kept in contact then the heat is transferred from the hot object to the cold object via different modes of heat transmission. Hence, the cold object starts getting hot
B) The transmission of heat from the hot object to the cold object will stop as the temperature of the two object becomes equal to each other.
Answer:
Explanation: simple kinematics
we suppose that initially vo= 0 so if the skier moves 4s :
vf = vo +at = 0 + 3*4 = 12 m/s
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Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Answer:
3a, 2b,4c,1d
Explanation:
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