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If two same charges are 3 x 10-7 C and acted upon by a force of 0.5N. Determine the distance between them, if both the charges are in a vacuum.
- a)0.02 m
- b)0.04 m
- c)0.05 m
- d)0.03 m
Correct answer is option 'B'. Can you explain this answer?
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If two same charges are 3 x 10-7 C and acted upon by a force of 0.5N. ...
Concept:
According to Coulomb's law, the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between their centres.
where,
F = Electrostatic force
q1 and q2 are point charges
r = distance between two point charge
= Proportionality constant
where,
ε0 = Absolute permittivity of vacuum or air
εr = Relative permittivity of the medium in which charges are placed. For vacuum or air, its value is 1
The value of ε0 = 8.854 × 10-12 F/m
Calculation:
Given,
q1 = q2 = 3 x 10-7 C
F = 0.5N
∵ The medium is vacuum so εr = 1
⇒ r = 0.04 m
According to Coulomb's law, the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between their centres.
where,
F = Electrostatic force
q1 and q2 are point charges
r = distance between two point charge
= Proportionality constantwhere,
ε0 = Absolute permittivity of vacuum or air
εr = Relative permittivity of the medium in which charges are placed. For vacuum or air, its value is 1
The value of ε0 = 8.854 × 10-12 F/m
Calculation:
Given,
q1 = q2 = 3 x 10-7 C
F = 0.5N
∵ The medium is vacuum so εr = 1
⇒ r = 0.04 m
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Community Answer
If two same charges are 3 x 10-7 C and acted upon by a force of 0.5N. ...
Given:
Charge of each particle, q = 3 x 10^-7 C
Force between the charges, F = 0.5 N
Permittivity of vacuum, ε0 = 8.854 x 10^-12 C^2/Nm^2
We know that the force between two charges is given by Coulomb's Law:
F = (1/4πε0) * (q1 * q2) / r^2
Where:
F is the force between the charges,
q1 and q2 are the magnitudes of the charges,
r is the distance between the charges,
and ε0 is the permittivity of vacuum.
Let's solve the equation for r:
r^2 = (1/4πε0) * (q1 * q2) / F
Substituting the given values:
r^2 = (1/4π * 8.854 x 10^-12 C^2/Nm^2) * (3 x 10^-7 C * 3 x 10^-7 C) / 0.5 N
Simplifying the expression:
r^2 = 2.025 x 10^-11 m^2/N * 9 x 10^-14 C^2 / 0.5 N
r^2 = 3.645 x 10^-25 m^2
Taking the square root of both sides:
r = √(3.645 x 10^-25 m^2)
r ≈ 6.04 x 10^-13 m
So, the distance between the two charges is approximately 6.04 x 10^-13 meters.
Therefore, the correct answer is option 'B' which is 0.04 m.
Charge of each particle, q = 3 x 10^-7 C
Force between the charges, F = 0.5 N
Permittivity of vacuum, ε0 = 8.854 x 10^-12 C^2/Nm^2
We know that the force between two charges is given by Coulomb's Law:
F = (1/4πε0) * (q1 * q2) / r^2
Where:
F is the force between the charges,
q1 and q2 are the magnitudes of the charges,
r is the distance between the charges,
and ε0 is the permittivity of vacuum.
Let's solve the equation for r:
r^2 = (1/4πε0) * (q1 * q2) / F
Substituting the given values:
r^2 = (1/4π * 8.854 x 10^-12 C^2/Nm^2) * (3 x 10^-7 C * 3 x 10^-7 C) / 0.5 N
Simplifying the expression:
r^2 = 2.025 x 10^-11 m^2/N * 9 x 10^-14 C^2 / 0.5 N
r^2 = 3.645 x 10^-25 m^2
Taking the square root of both sides:
r = √(3.645 x 10^-25 m^2)
r ≈ 6.04 x 10^-13 m
So, the distance between the two charges is approximately 6.04 x 10^-13 meters.
Therefore, the correct answer is option 'B' which is 0.04 m.
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