Question

A proton travels along the x-axis through an electric potential V=(250 V/m)x. Its speed is 3.5×10 5

m/s, as it passes the origin, moving in What is the proton's speed at x=1.0 m ? the +x-direction. Express your answer with the appropriate units.

Answer 1

To find the proton's speed at x=1.0 m, we can use the equation: Speed at x=1.0 m = Initial speed + (Electric potential at x=1.0 m * Distance). By substituting the given values, we can calculate the proton's speed as 350250 m/s.

Step-by-step explanation:

To find the proton's speed at a specific location, we need to consider the electric potential and the direction of motion. The electric potential, V, is given as V=(250 V/m)x, where x is the distance along the x-axis. Since the proton is moving in the positive x-direction, we can use the equation:

Speed at x=1.0 m = Initial speed + (Electric potential at x=1.0 m * Distance)

First, let's find the electric potential at x=1.0 m using the given electric potential function:

Electric potential at x=1.0 m = (250 V/m) * 1.0 m = 250 V

Now, we can calculate the proton's speed:

Speed at x=1.0 m = 3.5 × 10^5 m/s + (250 V * 1.0 m) = 3.5 × 10^5 m/s + 250 m/s = 350250 m/s

Answer 2

The proton's speed at x=1.0 m is approximately 2.4 x 10^5 m/s.

Step-by-step explanation:

To find the proton's speed at x=1.0 m, we can use the electric potential V=(250 V/m)x to calculate the change in potential energy. The change in potential energy is equal to the work done on the proton, which is given by the equation:

ΔPE = qΔV

Where q is the charge of the proton (1.6 x 10^-19 C) and ΔV is the change in electric potential. In this case, ΔV is given by ΔV = Vf - Vi, where Vf is the electric potential at x=1.0 m and Vi is the electric potential at x=0.

Since the electric potential is given by V=(250 V/m)x, we can substitute x=0 and x=1.0 into the equation to find the initial and final potentials. Plugging in the values, we get:

ΔV = (250 V/m)(1.0 m) - (250 V/m)(0 m) = 250 V

Finally, we can use the change in potential energy to find the change in kinetic energy:

ΔKE = -ΔPE = -qΔV = -(1.6 x 10^-19 C)(250 V) ≈ -4 x 10^-17 J

Since the proton's initial kinetic energy is given by KE = 0.5mv^2, where m is the mass of the proton (1.67 x 10^-27 kg) and v is its initial speed (3.5 x 10^5 m/s), we can solve for the final speed:

KE = 0.5mv^2 = 0.5(1.67 x 10^-27 kg)(3.5 x 10^5 m/s)^2 ≈ 1.84 x 10^-14 J

Setting the initial and final kinetic energies equal to each other, we get:

0.5mv^2 = -4 x 10^-17 J

Solving for v, we find:

v ≈ 2.4 x 10^5 m/s