two concentric conducting spherical shells have radii R1=0.145m and R2=0.207m the inner spheres with negligible speed, Assuming that the region between the spheres is a vacuum compate the speed with which the electron stricks the outer sphere."
To find the speed at which the electron strikes the outer sphere, we can use the principle of conservation of energy.
1. First, find the electric potential difference (ΔV) between the inner and outer spheres using the formula:
ΔV = k * (Q / R1 - Q / R2)
where k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), Q is the charge on the inner sphere, and R1 and R2 are the radii of the inner and outer spheres respectively.
2. Since the electron moves from the inner sphere to the outer sphere, it gains electric potential energy, which is equal to the kinetic energy it has when it reaches the outer sphere. Using the equation:
q * ΔV = 0.5 * m * v^2
where q is the charge of the electron (q = -1.6 x 10^-19 C), m is the mass of the electron (m = 9.1 x 10^-31 kg), and v is the final velocity of the electron.
3. Rearrange the equation to solve for v:
v = √(2 * q * ΔV / m)
Let's plug in the known values to calculate the speed at which the electron strikes the outer sphere:
ΔV = (8.99 x 10^9 Nm^2/C^2) * (Q / R1 - Q / R2)
= (8.99 x 10^9 Nm^2/C^2) * (Q/R1) * (1 - R1/R2)
Using the information given, we can proceed with the calculations. However, the value of the charge on the inner sphere (Q) is not provided. Please provide the charge on the inner sphere to proceed further in the calculations.
To calculate the speed with which the electron strikes the outer sphere, we can use the principle of conservation of energy. We will consider the electric potential energy and kinetic energy of the electron at different positions.
The electric potential energy of a charged particle in an electric field is given by the equation:
U = q * V
Where U is the potential energy, q is the charge of the particle, and V is the electric potential.
Inside a conductor, the electric potential is constant. Therefore, the initial potential energy of the electron at the inner sphere (R1) is equal to the potential energy at the outer sphere (R2). Since the electron is initially at rest, its kinetic energy is zero at R1.
At the outer sphere, the electron will have a kinetic energy and zero potential energy. Therefore, the total energy is conserved.
Mathematically, we have:
Initial energy at R1 = Final energy at R2
Since the electron has gained kinetic energy, we can write:
0 + 0 = 1/2 * m * v^2
Where m is the mass of the electron and v is its velocity.
Now, we can calculate the final velocity (v) by rearranging the equation:
1/2 * m * v^2 = 0
v^2 = 0
v = 0 m/s
Therefore, the speed with which the electron strikes the outer sphere is zero.