Two 2.5-cm-diameter disks face each other, 1.5 mm apart. they are charged to ± 14 nc. part a.
what is the electric field strength between the disks?
express your answer in newtons per coulomb.
part b.
a proton is shot from the negative disk toward the positive disk. what launch speed must the proton have to just barely reach the positive disk?
express your answer in meters per second.

Part A: The electric field strength between the discs =  

Part B: The launch speed of the proton =

Explanation:

Constant used:

Assumptions and Given values:

It is clear that the distance between the discs is much less than the diameter of each disc, therefore, the electric field in the region between the discs is nearly uniform.

According to the Gauss' law of electrostatics, the strength of the electric field due to charged disc at a point r distance away is given by

where, q is the charge on the disc and this expression is only valid if

r << (Diameter of the disc).

Assuming, the disc on left contains positive charge and the disc on right contains negative charge.

The strength of electric field between the discs due to the left disc is given by

The left disc is positively charged therefore the direction of this field is towards right.

Similarly, the strength of electric field between the discs due to the right disc is given by

The right disc is negatively charged therefore the direction of this field is also towards right.

Since, the electric fields due to both the discs are towards right therefore the strength of the net electric field between the discs due to both the discs is given by

We know,

According to Coulomb's law, the electrostatic force on the proton, between the discs, is given by

According to Newton's second law of motion,

It is the acceleration acquired by the proton between the discs.

The direction of the acceleration is from positive plate to negative plate.

When the proton moves from negative plate to positive plate its effective acceleration would be

Let the launch speed of the proton from the negative disc be u.

Since, the proton barely reach the positive plate therefore its speed when it touches the positive plate is 0, i.e., v = 0 m/s.

Using the equation,

It is the required launch speed.

see earlier answer

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