How does the electric force between two charged particles change if particle's charge is reduced by a factor of 3?
2 answers:
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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
<h2>Answer: 34.78 m/s</h2>
Explanation:
The momentum is given by the following equation:
(1)
Where:
is the mass of the object
is the velocity of the object
Finding the velocity from (1):
(2)
<u>Finally:</u>
>>>This is the velocity of the object