Question

Two 1.7-cm-diameter disks spaced 2.1 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.1×105 V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.5×107 m/s . What was the electron's speed as it left the negative plate?

Answer 1

The speed of the electron as it left the negative plate is 2.50x10^7 m/s.

Step-by-step explanation:

The electric field between the plates of a parallel-plate capacitor can be used to calculate the speed of an electron as it moves between the plates. In this case, we can use the given information to find the speed of the electron as it leaves the negative plate.

The electric field between the disks is given as 5.1×10^5 V/m.

To find the speed of the electron as it leaves the negative plate, we can use the equation for the electric field:

• E = V/d

Where:

• V is the voltage
• d is the separation between the plates.

Rearranging the equation, we have V = E*d. Plugging in the given values, we have V = (5.1×10^5 V/m)*(2.1 mm), which gives V = 1.071 V.

Since the electron starts from rest at the negative plate, the initial speed is 0 m/s. Using the equation for conservation of energy, we can find the final speed of the electron as it strikes the positive plate.

The equation is 1/2*m*v^2 = q*V, where m is the mass of the electron, v is the final speed, q is the charge of an electron, and V is the voltage between the plates.

Plugging in the known values, we have 1/2*(9.11x10^-31 kg)*(v^2) = -(1.6x10^-19 C)*(1.071 V).

Solving for v, we find v = 2.50x10^7 m/s. Therefore, the speed of the electron as it left the negative plate is 2.50x10^7 m/s.