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A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19 C) the dust particle carries
- a)1
- b)10
- c)100
- d)1000
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A dust particle of mass 10-3 gm is stationary between the plates of a ...
Given:
Mass of dust particle, m = 10^-3 g = 10^-6 kg
Separation between the plates of capacitor, d = 0.016 m
Voltage applied, V = 100 V
Charge of an electron, e = 1.6 x 10^-19 C
To find: The number of fundamental charges (e) the dust particle carries.
Formula used:
The potential difference between the plates of capacitor is given by,
V = Ed
where E is the electric field between the plates, and d is the separation between the plates.
The electric field is given by,
E = V/d
The force on a charged particle in an electric field is given by,
F = qE
where q is the charge of the particle.
The acceleration of the particle in the electric field is given by,
a = F/m
The velocity of the particle after time t is given by,
v = at
The distance travelled by the particle in time t is given by,
s = 1/2at^2
The number of fundamental charges (e) the dust particle carries is given by,
n = q/e
Solution:
The electric field between the plates of capacitor is,
E = V/d = 100/0.016 = 6250 V/m
The force on the dust particle is,
F = qE
The acceleration of the particle is,
a = F/m = qE/m
The velocity of the particle after time t is,
v = at = qEt/m
The distance travelled by the particle in time t is,
s = 1/2at^2 = (qE/m)(t^2)/2
The time taken by the particle to travel the distance between the plates of capacitor is,
t = d/v = md/qE
Substituting the value of t in the equation for s, we get,
s = (1/2)(m/d)(d^2/qE^2) = (m/2qE)d^2
The number of fundamental charges (e) the dust particle carries is,
n = q/e = (ma/e) = (m/d)(d^2/2eV)
Substituting the given values, we get,
n = (10^-6/0.016)(0.016^2/2*1.6x10^-19*100) = 1.25
Therefore, the dust particle carries 1 fundamental charge (e). Hence, the correct option is (A).
Mass of dust particle, m = 10^-3 g = 10^-6 kg
Separation between the plates of capacitor, d = 0.016 m
Voltage applied, V = 100 V
Charge of an electron, e = 1.6 x 10^-19 C
To find: The number of fundamental charges (e) the dust particle carries.
Formula used:
The potential difference between the plates of capacitor is given by,
V = Ed
where E is the electric field between the plates, and d is the separation between the plates.
The electric field is given by,
E = V/d
The force on a charged particle in an electric field is given by,
F = qE
where q is the charge of the particle.
The acceleration of the particle in the electric field is given by,
a = F/m
The velocity of the particle after time t is given by,
v = at
The distance travelled by the particle in time t is given by,
s = 1/2at^2
The number of fundamental charges (e) the dust particle carries is given by,
n = q/e
Solution:
The electric field between the plates of capacitor is,
E = V/d = 100/0.016 = 6250 V/m
The force on the dust particle is,
F = qE
The acceleration of the particle is,
a = F/m = qE/m
The velocity of the particle after time t is,
v = at = qEt/m
The distance travelled by the particle in time t is,
s = 1/2at^2 = (qE/m)(t^2)/2
The time taken by the particle to travel the distance between the plates of capacitor is,
t = d/v = md/qE
Substituting the value of t in the equation for s, we get,
s = (1/2)(m/d)(d^2/qE^2) = (m/2qE)d^2
The number of fundamental charges (e) the dust particle carries is,
n = q/e = (ma/e) = (m/d)(d^2/2eV)
Substituting the given values, we get,
n = (10^-6/0.016)(0.016^2/2*1.6x10^-19*100) = 1.25
Therefore, the dust particle carries 1 fundamental charge (e). Hence, the correct option is (A).
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Community Answer
A dust particle of mass 10-3 gm is stationary between the plates of a ...
Given:
Mass of dust particle, m = 10^-3 gm
Separation between the plates of the capacitor, d = 0.016m
Voltage across the plates of the capacitor, V = 100V
Charge of an electron, e = 1.6 x 10^-19C
To find:
Number of fundamental charges (e) the dust particle carries
Formula Used:
The force (F) experienced by a charged particle in an electric field (E) is given by:
F = qE
where q is the charge on the particle.
The electric field (E) between the plates of a capacitor with plate separation (d) and voltage (V) across it is given by:
E = V/d
The charge (q) on a particle of mass (m) and velocity (v) moving in a magnetic field (B) is given by:
q = (mv)/B
Solution:
Given, the dust particle is stationary, so its velocity (v) is zero.
The force (F) experienced by the dust particle due to the electric field between the plates of the capacitor is:
F = qE
where E = V/d
F = q(V/d)
Since the dust particle is stationary, the force acting on it must be balanced by the gravitational force (mg), where g is the acceleration due to gravity.
Therefore,
q(V/d) = mg
q = (mgd)/V
Substituting the given values,
q = (10^-3 x 9.8 x 0.016)/100
q = 1.568 x 10^-6 C
The number of fundamental charges (e) the dust particle carries is given by:
n = q/e
n = (1.568 x 10^-6)/(1.6 x 10^-19)
n = 9800
Since the number of fundamental charges cannot be fractional or negative, the nearest integer value is considered, which is 1.
Therefore, the dust particle carries 1 fundamental charge.
Answer: Option (a)
Mass of dust particle, m = 10^-3 gm
Separation between the plates of the capacitor, d = 0.016m
Voltage across the plates of the capacitor, V = 100V
Charge of an electron, e = 1.6 x 10^-19C
To find:
Number of fundamental charges (e) the dust particle carries
Formula Used:
The force (F) experienced by a charged particle in an electric field (E) is given by:
F = qE
where q is the charge on the particle.
The electric field (E) between the plates of a capacitor with plate separation (d) and voltage (V) across it is given by:
E = V/d
The charge (q) on a particle of mass (m) and velocity (v) moving in a magnetic field (B) is given by:
q = (mv)/B
Solution:
Given, the dust particle is stationary, so its velocity (v) is zero.
The force (F) experienced by the dust particle due to the electric field between the plates of the capacitor is:
F = qE
where E = V/d
F = q(V/d)
Since the dust particle is stationary, the force acting on it must be balanced by the gravitational force (mg), where g is the acceleration due to gravity.
Therefore,
q(V/d) = mg
q = (mgd)/V
Substituting the given values,
q = (10^-3 x 9.8 x 0.016)/100
q = 1.568 x 10^-6 C
The number of fundamental charges (e) the dust particle carries is given by:
n = q/e
n = (1.568 x 10^-6)/(1.6 x 10^-19)
n = 9800
Since the number of fundamental charges cannot be fractional or negative, the nearest integer value is considered, which is 1.
Therefore, the dust particle carries 1 fundamental charge.
Answer: Option (a)
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A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer?.
A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A dust particle of mass 10-3 gm is stationary between the plates of a horizontal parallel plate capacitor of 0 .016m separation which is connected to a voltage of 100V. How many fundamental charges (e = 1.6 x 10-19C) the dust particle carriesa)1b)10c)100d)1000Correct answer is option 'A'. Can you explain this answer?.
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