The electric field between plates (A) is zero. As the beam...

The electric field between plates (A) is zero. As the beam exits the space between plates (B), it has been deflected $2.0 \mathrm{mm}$ downward $(\Delta y=-2.0 \mathrm{mm}) .$ What is the electric field between plates (B)?

Key Concepts

Uniform Electric Field

A uniform electric field is one that has the same magnitude and direction at every point in a specified region. In contexts such as the space between two parallel plates, this results in a constant force on a charged particle entering the field, simplifying the analysis of its motion.

Force on a Charged Particle

When a charged particle is placed in an electric field, it experiences a force directly proportional to both its charge and the strength of the field (F = qE). This force is what causes the particle to accelerate and ultimately deflect from its original path.

Kinematics under Constant Acceleration

A charged particle moving through a uniform electric field experiences constant acceleration. Using kinematic equations, the displacement of the particle in the direction of the force can be related to the acceleration and the time spent in the field. This analysis is key to determining the electric field strength from the observed deflection.

Transcript

00:01 High in the given problem the set of plates which is marked as b is horizontal like this so the electric field produced between them will be vertical and as the electrons which were initially moving in the direction of x x x x x.

00:35 So as the electrons are deflecting in a vertically downward direction means negative y -axis and as electrons are deviated towards the positively charged plate so we can say this lower plate is positively charged and upper plate is negatively charged that means electric field is in a vertically upward direction whose magnitude here is missing, which we have to find.

01:33 So as electrons are deflecting in a vertically downward direction, so the direction of electric field is upward along positive y -axis.

02:01 Now, the deflection suffered by these electrons along y -axis is given as delta y is equal to 2 .0 millimeter in magnitude.

02:18 So we are ignoring the negative sign associated with it.

02:23 Initial vertical velocity when electrons enter into this electric field, they are moving purely along z -axis.

02:31 So this is viz so initial vertical velocity of the electrons is zero and the electrons then will go like this they will be deflected and this is the vertical deflection delta y now initial vertical velocity of the electrons is 0 means we can say v i y is equal to 0 and the motion of electrons along z -axis is uniform as they are experiencing force only along y -axis means only due to the electric field which is along y -axis so there will be no other force so there will be no other acceleration so the motion will be uniform along x xx is and the distance move by the electron within the electric field is equal to the size of the side of the plate which is given as 2 .50 centimeter therefore time taken by the electron within the electric field will be equal to time is equal to distance means let it be z because it is covered in that z direction that's why the distance means the length of the plate we are saying it to be z divided by v i z initially speed along z axis which is given as here this is 2 .50 centimeter or 2 .50 into 10 dash per minus 2 meter divided by the speed along x x x which is given as 8 .4 into 10 dash 6 meter per second so this time comes out to be 2 .98 into 10 to the power minus 9 second...

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