(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².
(b) The speed of the electron after the given time is 4.78 x 10⁵ m/s.
<h3>
Acceleration of the electron</h3>
The acceleration of the electron is calculated as follows;
F = qE
ma = qE
a = qE/m
a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)
a = 5.62 x 10¹³ m/s²
<h3>Speed of the electron</h3>
v = at
v = 5.62 x 10¹³ m/s² x 8.50 x 10⁻⁹ s
v = 4.78 x 10⁵ m/s
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a. all systems work together to stabilize the body
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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The value of 'g' is not affected by rotation at any place on Earth.
Im pretty sure its the first option
