Answer:
U_eq = 1.99 * 10^(-10) J
Explanation:
Given:
Plate Area = 10 cm^2
d = 0.01 m
k_dielectric = 3
k_air = 1
V = 15 V
e_o = 8.85 * 10 ^-12 C^2 / N .m
Equations used:
U = 0.5 C*V^2 .... Eq 1
C = e_o * k*A /d .... Eq 2
U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3
For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:
U_electric = 0.5 e_o * k_1*A*V^2 /2*d
U_air = 0.5 e_o * k_2*A*V^2 /2*d
The total Energy is:
U_eq = U_electric + U_air
U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d
U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d
Plug the given values:
U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01
U_eq = 1.99 * 10^(-10) J
Answer:
Fast moving electrons produce larger current because if electrons move faster, the rate of flow of charge in a given volume will be more
Explanation:
Assuming that there are no collisions between the electrons
Fast moving electrons produce larger currents because the rate of flow of charge which is the rate of change of amount of charge in a given volume will be more as the speed of the electrons is more
<h3>As the current is defined as the rate of flow of charge and by the formula it is represented as </h3><h3>I = dQ ÷ dt</h3>
where I is the current flowing through it
dQ is the change in amount of charge
dt is the change in time
In case of fast-moving electrons the term dQ ÷ dt will be more and therefore the current will be more
∴ Larger current is produced by fast moving electrons
Answer:
asi que sobre todo lol siento soy mexican y agora no se Como escribir en ingles
soi aamso jodidamente lol siento que me enfrie want thig ror you
Explanation:
asi que sobre todo lol siento soy mexican y agora no se Como escribir en ingles
soi aamso jodidamente lol siento que me enfrie want thig ror you
Answer: Yes
The capacitor in an RC circuit (Resistor-Capacitance circuit) creates a time delay that decays exponentially. Therefore when the voltage to the bulb is removed, it takes a long time for the capacitor to completely discharge.
Although the brightness of the bulb will diminish as the capacitor discharges, the bulb will still be lit until the capacitor discharges to a voltage that is too low to sustain visible light (to the human eye) from the bulb.
