Hefei v2 photography

Let the charge of the ball be q and the mass be m, and the gravity work above the plate is a:12mv21= mgl1.

When passing through the electric field, the electric field force and gravity do work, and the result is:12mv22 = mg (l+13l)+QE13l②.

Meaning: V 1: V2 = 3: 5? ③

Simultaneous solution: mg=qE ④.

The direction of the electric field is downward.

After re-release, after the ball enters the composite field, because of the reverse electric field, the direction of electric field force is upward, qE=mg, so the particle moves in a uniform circle.

Lorentz force provides centripetal force: QV 1B = MV2 1R? ⑤

In order to make the ball fly out of the field without colliding with the electrode plate, it should meet the requirements of 14L < R < 13L6.

6 e2gl < b < 8 e2gl ⑦ are obtained by simultaneous solution.

Answer: After the ball enters the space between the two plates from the small hole of the A plate, it can fly out of the space between the two plates without colliding with the polar plate. The range of magnetic induction intensity b is 6e2gl < b.