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A charged particle placed in an electric field falls from rest through a distance d in time t. If the charge on the particle is doubled, the time of fall through the same distance will be
- a)2t
- b)t
- c)t/√2
- d)t/2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A charged particle placed in an electric field falls from rest through...
√2t
d = 1/2at^2, where a is the acceleration of the particle in the electric field.
We can also write a as a = F/m, where F is the electric force on the particle and m is its mass.
Therefore, d = 1/2(F/m)t^2, or F = 2md/t^2.
Now, if we double the charge on the particle, the electric force on it becomes F' = 2qE, where q is the charge and E is the electric field strength.
So, we have F' = 2qE = 2(2q/2)E = 2q'E, where q' is the new charge on the particle.
Therefore, the new acceleration of the particle is a' = F'/m = (2q'E)/m.
To find the time of fall through the same distance, we use d = 1/2a't^2, which gives us:
t = √(2d/a') = √(2md/q'E) = √(2t^2(q/m)(d/E)).
Since q/m and d/E are constants for the same particle and electric field, we can see that the time of fall is proportional to the square root of the charge on the particle.
Therefore, if we double the charge on the particle, the time of fall through the same distance will be t/√2. Answer: (c) t/√2.
d = 1/2at^2, where a is the acceleration of the particle in the electric field.
We can also write a as a = F/m, where F is the electric force on the particle and m is its mass.
Therefore, d = 1/2(F/m)t^2, or F = 2md/t^2.
Now, if we double the charge on the particle, the electric force on it becomes F' = 2qE, where q is the charge and E is the electric field strength.
So, we have F' = 2qE = 2(2q/2)E = 2q'E, where q' is the new charge on the particle.
Therefore, the new acceleration of the particle is a' = F'/m = (2q'E)/m.
To find the time of fall through the same distance, we use d = 1/2a't^2, which gives us:
t = √(2d/a') = √(2md/q'E) = √(2t^2(q/m)(d/E)).
Since q/m and d/E are constants for the same particle and electric field, we can see that the time of fall is proportional to the square root of the charge on the particle.
Therefore, if we double the charge on the particle, the time of fall through the same distance will be t/√2. Answer: (c) t/√2.
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Community Answer
A charged particle placed in an electric field falls from rest through...
On doubling the charge, to force doubles as
F = q E
also F = m a
∴ a doubles
F = q E
also F = m a
∴ a doubles
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