Question

Find the electrostatic force of interaction between two halves
of a spherical conductor of radius R carrying a charge Q.​

Answer 1

Answer:

See explanation

Explanation:

Take an elementary area dA containing a charge dq on

the elementary ring of radius r. The force acting on dq is

dF=σ22ε0dA (as derived earlier)

The net force on the ring is

Fr∈g=∫dFsinθ=∫σ22ε0dAsinθ

=σ22ε0sinθ∫dA=σ22ε0sinθ(Ar∈g)

where Ar∈g=2πrdthη=2π(Rcosθ)Rdθ=2πR2cosθdthη

or Fr∈g=σ2πR2ε0 sin theta cos theta dtheta...........(i)F_(hemisphere) = int F_(ring)..........(ii)Substitut∈gF_(ring)omEq.(ii),we≥tF_(hemisphere) = (sigma^2piR^2)/epsilon_0∈t_(0)^(+pi//2)sinθcosθ,whereσ= Q/(4piR^2)= (Q^2)/(32piepsilon_0R)`

Answer 2

Answer:

The electrostatic force of interaction between two halves of a spherical conductor of radius R carrying a charge Q. is $\frac{Q^{2} }{32pieR^{2}e_{o} }$

Explanation:

F=p*A

Where p is the unit of pressure in pascals,

The newton (f) unit of force equals one pascal (1 N/m 2).

Electrostatic pressure= μ$^{2}$/$2e_{o} }$

μ=$\frac{Q }{4pieR^{2}}$

μ$^{2}$=$\frac{Q^{2} }{16(pie)^{2} R^{4}}$

F=$\frac{Q^{2} }{16(pie)^{2} R^{4}2e_{o} }*(pie)R^{2}$

F=$\frac{Q^{2} }{32pieR^{2}e_{o} }$

Electrostatic force:

• The study of electric charges in a resting state is called electrostatics.
• Some materials, like amber, have been known to capture light particles after rubbing since classical times.
• Thus, the term "electricity" derives from the Greek word for amber, "v."
• The attraction or repulsion between particles that electrostatic forces produce is due to their electric charges In honor of French physicist Charles-Augustin de Coulomb, who first characterized the force in 1785, this force is also known as the Coulomb force or Coulomb interaction.

SPJ3