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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.
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
q_{1} and q_{2} are point charges
r = distance between two point charge
= Proportionality constant
where,
ε_{0} = Absolute permittivity of vacuum or air
ε_{r} = 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,
q_{1} = q_{2} = 3 x 10^{7} C
F = 0.5N
∵ The medium is vacuum so ε_{r} = 1
⇒ r = 0.04 m
It states that the force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.
F ∝ q_{1} ∝ q_{2}
Where
also called Coulomb's constant or electrostatic force constant.
Here ϵ_{0} = Permittivity of free space.
The value of K depends on the nature of medium between the two charges and the system of units chosen.
The electric field of a point charge is a linear function of the value of the charge. It is the principle applied to an electric field and states that the total resultant field at a point is the vector sum of the individual components of the field at a point. Identify the principle mentioned in the context.
CONCEPT:
Superposition Principle:
Coulomb's law in Electrostatics:
Where K = constant called electrostatic force constant.
Two charges of + 4 μC and 16 μC are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 μC be placed from + 4 μC so that net force exerts on it will be zero?
CONCEPT:
Coulomb’s law: When two charged particles of charges q_{1} and q_{2} are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
Force (F) ∝ q_{1} × q_{2}
Where K is a constant = 9 × 10^{9} Nm^{2}/C^{2}
Calculations:
Consider new charge + 4μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from 16 μC charge.
Let,
q_{A} = + 4 μC at point A
q_{B} =  16 μC at point B
q_{C} = + 6 μC at point C
Since, net force on charge q_{C} will zero.
∴ F_{CA} = F_{CB}
From above concept,
4d^{2} = (0.6 + d)^{2}
4d^{2} = 0.36 + d^{2} + 1.2d
3d^{2}  1.2d  0.36 = 0
d_{1} =  0.2 m
d_{2} = + 0.6 m
Hence, according to option + 0.6 m distance should a third charge of + 6 μC be placed from + 4 μC so that no force exerts on it will be zero.
Coulomb’s Law: It states that the force between two charged particles will be directly proportional to the product of the quantity of the two charges.
F ∝ Q_{1}Q_{2},
Secondly, it is also inversely proportional to the square of the distance between them, i.e.
F ∝ 1/r2
∴ The force between two charges Q_{1} and Q_{2} is given by:
where k is the proportionality constant and is equal to 9 × 10^{9} Nm^{2}/C^{2}.
Consider the two point particles separated by a distance 'd' have charges Q_{1} and Q_{2} respectively. Particle Q_{2} experiences an electrostatic force of 20 mN due to particle Q_{1}. If the charges of both particles are doubled and distance between them is also doubled, what is the magnitude of the electrostatic force between them?
Concept:
Force (F) ∝ q_{1} × q_{2}
Where K is a constant = 9 × 10^{9} Nm^{2}/C^{2}
Case 1:
Given that charges are: Q_{1} and Q_{00}
Distance = d
Force (F_{1}) = 20 mN
Case 2:
The charges of both particles are doubled and distance between them is also doubled.
Given that charges are: 2Q_{1} and 2Q_{2}
Distance = 2d
Therefore, the force remains same and it is equal to 20 mN
False statement among the following statements regarding Coulombs law is:
CONCEPT:
Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
Force (F) ∝ q_{1} × q_{2}
Where K is a constant = 9 × 10^{9} Nm^{2}/C^{2}
According to which law, the direction of induced emf in the choke of a tube light will be such that it will try to oppose the fall of the current in the circuit?
Lenz’s Law:
A charge Q_{2} = 8.854 × 10^{9} C is located in vacuum at P_{2} (2, 3, 1). The force on Q_{2} due to a charge Q_{1} = 4π × 10^{3} C at P_{1} (2, 2, 1) is:
(Note: All the coordinates are measured in meters a_{x}, a_{y}, and a_{z} are unit vectors in X, Y, and Z direction respectively.)
Concept:
The force acting on a point charge because of another point charge is given as:
(Where â_{r} is a unit vector in the direction of the force)
Calculation:
Given Q_{1} = 4π × 10^{3} C
Q_{2} = 8.854 × 10^{9} C
Since, we are asked to find the force on Q_{2} due to Q_{1}, the direction of the force
= â_{r}
So, the Force on Q_{2} due to Q_{1} will be:
= a_{y} N
The law, governing the force between electric charges is known as:
CONCEPT:
Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
Force (F) ∝ q_{1} × q_{2}
Where K is a constant = 9 × 10^{9} Nm^{2}/C^{2}
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