The plates create uniform electric field to the right in the picture, with magnitude `[V_0 - (-V_0)]// d = 2V_0 //d`. Assume the ball swings a small distance x to the right. It moves to a place where the voltage created by the plates is lower by
`- E x = -(2 V_0)/d x`
Its ground connection maintains it at `V = 0` by allowing charge q to flow from ground onto the ball, where `- (2V_0 x)/d + (k_e q)/R = 0`
`q = (2 V_0 x R)/(k_e d)`
Then the ball feels electric force
`F = q E = (4 V_0^2 x R)/(k_e d^2)` to the right.
for equilibrium, this must be balanced by the horizontal component of string tension according to `T cos theta = mg`
`T sin theta = (4 V_0^2 x R)/(k_e d^2) tan theta = (4 V_0^2 x R)/(k_e d^2 mg) = (x)/(L)` for small x
Then `V_0 = ((k_e d^2 m g)/(4 R L))^(1//2)`
If `V_0` is less than this value, the only equilibrium position of the ball is hanging straight down. If `V_0` exceeds this value, the ball will saving over to one plate or the other.