In the device shown in the figure, there is a small ball (which can be regarded as a particle) with a mass of m and a charge of q in the electric field of a parallel plate, which is tied with a thin l

In the device shown in the figure, there is a small ball (which can be regarded as a particle) with a mass of m and a charge of q in the electric field of a parallel plate, which is tied with a thin line with a length of L and then located in the electric field. Your picture was not sent because of its low level, so some things are unknown. The following is my analysis.

(1) From the meaning of the title, two parallel plates are vertical and the direction of uniform electric field is horizontal.

If the direction of electric field is left and the ball is left, the ball is positively charged; If the ball leans to the right, it is negatively charged.

If the direction of electric field is right and the ball is right, the ball is positively charged; If the ball leans to the left, it is negatively charged.

(2) Analyze the force on the ball: gravity mg, rope tension t, electric field force f, and the resultant force is 0.

F = QE = mg * tan θ comes from triangle knowledge.

The electric field intensity is e = mg * tanθ/q.

And because e = u/d, the potential difference between the two plates is u = ed = d mg * tan θ/q.

(3) Let the speed of the ball when it reaches the equilibrium position be V..

QE * l * sin θ-mg * l * (1-cos θ) = m * v2/2 comes from the kinetic energy theorem.

Substitute E = mg * tan θ/q into the above formula to obtain.

The required speed v = root sign [2gl (tan θ * sin θ- 1+cos θ)]

(4) Setting the initial velocity of the ball at the equilibrium position as V0 can make the ball do circular motion just in the vertical plane.

Combine gravity and electric field force to get an equivalent gravity, and G effect = radical sign [(QE) 2+(mg) 2] = mg/cos θ.

The equilibrium position is the equivalent "lowest point", another point with the same diameter as the equilibrium position is the equivalent "highest point", and the speed at the equivalent "highest point" is V 1.

Then the -g effect * 2l = (m * v 1 2/2)-(m * v0 2/2) is obtained from the kinetic energy theorem.

At the equivalent "highest point", there is an F direction = G effect = m * V12/L.

Substituting G effect = mg/cos θ into the above two formulas, the initial velocity obtained after finishing is

V 0 = radical sign (5gL/cosθ)