Construction. It consists of four similar hollow metallic quadrants AA and BB supported separately on amber insulting stands. The opposite pairs of quadrants AA and BB are connected together by fine copper wires. A paddle-shaped aluminum needle with two wings CC is suspended symmetrically between the four quadrants by means of a torsion fibre made of phosphor bronze.

Working. The needle N is usually kept at a constant high potential V_n by connecting it to the positive pole of a H.T. battery whose negative is earthed. When the two pairs of quadrants, AA and BB, are charged to the same potential, the needle rests symmetrically between them. If they are given different potentials, say V_a and V_b such that V_a > V_b, then the needle deflects from A quadrants to B quadrants.

Theory. Let r be the radius of the needle. With the deflection of the needle through an angle θ,  a surface area 1r²θ/2 of the needle is transferred from the quadrants AA to BB. Since the needle has two arms and two faces, the total area transferred from AA to BB is given by
        A = 4 X 1r²θ/2 = 2r²θ.

The shift in area causes in increases in the capacitance of the B – N capacitor given by
        δC = ε₀A = 2r²θε₀
________d           d

where d is the distance between the surface of the needle and the top or bottom of the quadrants.

The capacitance of the A – N capacitor decreases by the same amount.
Therefore, the increase in the energy of the B – N capacitor
= ½ X δC X (P.D.)²
= ½ X 2r²θε₀/d X (V_n - V_b)² = r²θε₀/d (V_n – V_b)²

Similalry, the decrease in the energy of the A – N capacitors
 = r2θε0/d (Vn - Va)2

. : net increase in the energy of the system.
 = r2θε0/d [(Vn - Vb)2  - (Vn – Va)2].

In addition to this net gain in electrical potential energy, work has also to be done in twisting the suspension fibre by an angle θ. It is given by ½ cθ² where c is the restoring couple per unit twist of the suspension fibre.
. :    total energy gain
= ½ cθ² + r²θε₀/d [(Vn - Vb)² – (V_n – V_a)²]._________…. (i)

As the capacitance of B – N capacitor increases by δC, it draws a charge δC X (V_n - V_b) from the source.
Energy drawn from the source at a constant p.d. ((V_n - V_b) from the source.
= charge X potential difference
= {δC X (V_n - V_b)} X (V_n - V_b)
= 2r₂θε₀/d X (V_n - V_b)².

Similarly, the A – N capacitor, whose capacitance decreases by δC, restores to the source an amount of energy.
= 2r²θε₀/d X (V_n - Vb)².
. : net energy drawn from the source
= 2r²θε⁰/d [(V_n - V_b)² – (V_n - V_a)²]. __________…. (ii)

This must be equal to the total energy gained by the electrometer.
Hence equating (i) and (ii), we get
½ cθ² + r²θε₀/d [(V_n - V_b)² – (V_n – V_a)²] = = 2r²θε₀/d [(V_n - V_b)² – (V_n - V_a)²]
or ½ cθ² = + r²θε₀/d [(V_n - V_b)² – (V_n – V_a)²] = 2r²θε₀/d (V_a - V_b) [ V_n - V_a + V_b/2]
or θ = 4r²θε₀/cd (V_a - V_b) [ V_n - V_a + V_b/2]
. : θ = k (V_a - V_b) [ V_n - V_a + V_b/2],__________  …..  (iii)
Where k 4r²θε₀/cd

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