Each of the protons in a particle beam has a kinetic energy of 3.05 10-15 J. What are the magnitude and direction of the electric field that will stop these protons in a distance of 1.15 m?
Voltage * charge = energy
So you know what voltage you need.
E * distance = Voltage
so you know distance = 1.15 so solve for E
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To find the magnitude and direction of the electric field that will stop the protons, we can use the following steps:
1. Determine the charge of a proton. The charge of a proton is equal to the elementary charge, which is approximately 1.6 x 10^(-19) C.
2. Calculate the velocity of the protons using the kinetic energy given. The kinetic energy (KE) of a particle is given by the equation KE = (1/2)mv^2, where m is the mass of the particle and v is its velocity. Since we know the kinetic energy and the mass of a proton is approximately 1.67 x 10^(-27) kg, we can solve for v.
KE = (1/2)mv^2
3.05 x 10^(-15) J = (1/2)(1.67 x 10^(-27) kg)v^2
v^2 = (2 * 3.05 x 10^(-15) J) / (1.67 x 10^(-27) kg)
v^2 ≈ 3.663 x 10^12 m^2/s^2
v ≈ √(3.663 x 10^12 m^2/s^2)
v ≈ 6.05 x 10^6 m/s
3. Calculate the time it takes for the protons to stop. Distance (d) is equal to velocity (v) multiplied by time (t). Identify that the distance is given as 1.15 m. We can rearrange the equation to solve for time.
d = vt
1.15 m = (6.05 x 10^6 m/s)t
t ≈ (1.15 m) / (6.05 x 10^6 m/s)
t ≈ 1.898 x 10^(-7) s
4. Use the formula for the force experienced by a charged particle in an electric field: F = qE, where F is the force, q is the charge, and E is the electric field strength. We know the force required to stop the protons is equal to the change in momentum, which is given by Δp = mv, where m is the mass and v is the velocity. We can rearrange the equation to solve for the electric field (E).
Δp = FΔt
mv = qEΔt
E = (mv) / (qΔt)
E = (1.67 x 10^(-27) kg)(6.05 x 10^6 m/s) / [(1.6 x 10^(-19) C)(1.898 x 10^(-7) s)]
E ≈ 5.89 x 10^4 N/C
Therefore, the magnitude of the electric field required to stop the protons is approximately 5.89 x 10^4 N/C in the opposite direction of their motion.