When in air , for each ball , T cos\ heta = Density \ imes volume \ imes g = dVg ( Vertical component of tension is balanced by weight ) T sin\ heta = \\dfrac{q^{2} }{4\\pi \\epsilon_0 ( 2l sin \ heta) ^{2} } ( The horizontal component of tension is balanced by the Electrostatic Force ) tan\\ \ heta=\\dfrac{\\dfrac{q^{2}}{4\\pi \\epsilon_0 (2l\\ sin\ heta)^2}}{dVg} ----------- (1)Similarly ,when immersed in liquid , tan \ heta = \\dfrac{ \\dfrac{ q^{2} }{4\\pi K \\epsilon_0 (2l sin \ heta) ^{2} }}{Vg ( d-d_{1} )} ----------(2) ( Buoyant force, gravity and electrostatic force ) From (1) and (2) K(d-d_{1}) = d Hence , d = \\dfrac{K d_{1} }{K-1}