Two concentric conducting spherical shell have raddii R1=0.145m and R2 =0.207m. The inner sphere with negligible speed. Assuming that the region between the sphere is a vacuum, compute the speed with which the electron strikes the outer sphere.

Quite a lot is missing here.

2.0000m/s

To find the speed with which the electron strikes the outer sphere, we can apply the principle of conservation of energy. We'll consider the initial potential energy of the electron at the inner sphere and its final kinetic energy at the outer sphere.

The potential energy of the electron at the inner sphere can be calculated using the formula for potential energy of a charged particle in an electric field:

PE_initial = q * V_initial

Where q is the charge on the electron and V_initial is the potential difference between the inner and outer spheres. Since the inner sphere is grounded (assumed to have negligible potential), V_initial is equal to the potential at the outer sphere.

The potential at the outer sphere (V_final) can be calculated using the formula for the electric potential due to a charged sphere:

V_final = k * (Q_final / R2)

Where k is the Coulomb's constant, Q_final is the charge on the outer sphere, and R2 is its radius.

The charge on the electron (q) is known (-e, where e is the elementary charge).

Now, we can equate the initial potential energy to the final kinetic energy:

PE_initial = KE_final

This gives us:

-q * V_final = (1/2) * m * v_final^2

Where m is the mass of the electron and v_final is its final velocity.

Solving this equation for v_final, we can find the speed with which the electron strikes the outer shell.

Note: To solve this problem, we need the values for the charge on the outer sphere (Q_final), mass of the electron (m), and Coulomb's constant (k).