Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Energy can change form, but the total amount of energy stays the same.
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
The charges are the same in absolute value, so the change of potential energy is the same. That means that the change in kinetic energy is also the same. Then:
1 = Ke/Kp = m_e *v_e^2 / m_p * v_p^2, or
v_e/v_p = sqrt( m_p/m_e),
So the speed of the electron will be sqrt( m_p/m_e) times greater than the speed of the proton