A positively charged particle of certain mass may be suspended (at rest) in electric field of suitable strength if the field is
1 answer:
If the electric field is directed d) upward
Explanation:
In order for a positively charged particle to be suspended at rest, the electric force due to the electric field must be
1) equal in magnitude, and
2) opposite in direction
to the force of gravity acting on the particle. Since the force of gravity acts downward, the electric force must be upward. The electric force on a charged particle is given by
where
q is the charge
E is the electric field
And the direction of the force is:
- The same as the electric field for a positive charge
- Opposite to the direction of the field for a negative charge
Here we have a positive charge, therefore the direction of the electric field must be the same as the direction of the force, therefore upward.
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Answer:
t = 23.9nS
Explanation:
given :
Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m
distance d= 1mm=> 0.001
resistor R= 975 ohm
Capacitance can be calculated through the following formula,
C = (ε0 x A )/d
C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001
C = 17.7 x 10^-12 (pico 'p' = 10^-12)
C = 17.7pF
the voltage between two plates is related to time, There we use the following formula of the final voltage
Vc = Vx (1-e^-(t/CR))
75 = 100 x (1-e^-(t/CR))
75/100 = (1-e^-(t/CR))
.75 = (1-e^-(t/CR))
.75 -1 = -e^-(t/CR)
-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)
e^-(t/CR) = 0.25
in order to remove the exponent, take logs on both sides
-t/CR = ln (0.25)
t/CR = -ln(0.25)
t = -CR x ln (0.25)
t = -(17.7 x 10^-12 x 975) x (-1.38629)
t = 23.9 x
t = 23.9ns
Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts