Two charges of equal magnitude q are placed in air at a distance of 2a?

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Explanation of Two Charges Placed in Air

When two charges of equal magnitude q are placed in air at a distance of 2a, the following happens:

Electrostatic Force

The two charges experience an electrostatic force of attraction towards each other due to the opposite charges. This force is given by Coulomb's law:

F = (1/4πε₀) * (q²/2a²)

where F is the force,

ε₀ is the permittivity of free space,

q is the magnitude of the charges, and

a is the distance between the charges.

Charge Distribution

The charges will redistribute themselves to minimize their potential energy due to electrostatic forces. The charges will move towards each other until they reach equilibrium at a distance where the electrostatic force is balanced by the repulsion due to their proximity. This distance is called the equilibrium separation distance.

Electric Field

The charges create an electric field in the space around them. The electric field is given by:

E = (1/4πε₀) * (q/2a²)

where E is the electric field,

ε₀ is the permittivity of free space,

q is the magnitude of the charges, and

a is the distance between the charges.

The electric field is a vector field that points in the direction of the force that a positive test charge would experience if placed in the field. The electric field is strongest near the charges and decreases with distance.

Answer

Dipole bn gya na
q × 2a =p

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