# Electrostatic question

Consider a see-saw, with central pivot, made from a long rod of insulating material. The left end of the see-saw supports a conducting sphere with a charge Q = 5.0 × 10−6 C, which experiences an electrostatic force from an identical sphere 10 cm below carrying a charge +Q. The right end of the see-saw supports another conducting sphere with a charge of +3Q, which is 10 cm above a sphere carrying a charge +4Q A block is placed on the right end of the see-saw to balance the system and to keep the rod horizontal. The mass of the spheres and rod can be neglected in the calculations.

(1)Calculate the mass of the block required to balance the see-saw.

The force on the left side is

$F1 =Q^2/(4 \pi \epsilon_0*r^2)$ where $\epsilon_0$ is the dielectric permittivity of the free vacuum (air) $= 8.84*10^{-12} F/m$ and r is the distance between the charges.

$F1 =(5*10^{-6})^2*9*10^9/0.1/0.1 =22.5 N$

The force on the right side is

$F2 =12*Q^2/(4 \pi \epsilon_0*r^2) =12*(5*10^{-6})^2*9*10^9/0.1/0.1 =12*F1 =270 N$

The weight of block is simply

$G =F2-F1 =270-22.5 =247.5 N$

The mass of the block is

$m= G/g =247.5/9.81=25.23 Kg$