Two identical metal balls are suspended on threads of equal length, fixed at one point. When the balls were given charges

Two identical metal balls are suspended on threads of equal length, fixed at one point. When the balls were given charges of the same magnitude and sign, the filaments parted at a certain angle. What should be the dielectric constant of a liquid dielectric so that when this system is immersed in it, the angle of divergence of the threads does not change? The ratio of the density of the material of the balls to the density of the liquid dielectric is 3.

T- thread tension
F- electric force
F- Archimedean force
The axes go through the ball Oh-right, Oh-up
Let’s compose Newton’s II law for the position of balls in air (vector):
T + F + mg = 0
Оx: F = T sin a
Оy: mg = T cos a
we divide these two expressions into each other, instead of F we substitute the formula – k * q ^ 2 / r ^ 2, we get:
tg a = k * q ^ 2 / r ^ 2 * mg
Let’s compose Newton’s II law for the position of balls in a liquid (vector):
T + F + mg + Fa = 0
ox: F = T sin a
oy: mg-Fa = T cos a
we divide these two expressions into each other, instead of F we substitute the formula – k * q ^ 2 / r ^ 2 * E (dielectric constant), instead of Fa – ρzh * g * m / ρt, (Vt = m / ρt) we get:
tan a = k * q ^ 2 / r ^ 2 * (mg-1 / 3mg) * E
Since the angle has not changed, the tangents are equal, we can equate the expressions and reduce, we get:
1 / mg = 3/2 * mg * E => E = 1.5
Answer: 1.5

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