Answer:
Explanation:
<u>Energy Stored on a Capacitor</u>
This energy is stored in the electric field present when a voltage V is applied to a capacitor C. It can be computed by the formula
We can compute the energy by knowing the power dissipated in the flashlamp is P=8 W in a time of
Solving the first equation for C
Answer:
Explanation:
Area of electrodes, A = 2 cm x 2 cm = 4 cm²
Separation between electrodes, d = 1 mm
Voltage, V = 9 V
(a)
Let C is the capacitance between the electrodes
C = 3.54 x 10^-12 F
Let q be the charge on each of the electrode
q = C x V
q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C
(b)
As, the battery is disconnected the charge on the electrodes remains same.
(c)
As the battery is connected the voltage is same.
capacitance is change.
As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C