Question

A proton is released from rest at point A in a uniform electric field E that has a magnitude of 8.0 x 10^4 v/m. The proton undergoes a displacement d=0.50m to point B in the direction of E. Calculate the speed of the proton after completing the displacement.

Answer 1

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Answer on Question #68150 Physics / Other

A proton is released from rest at point A in a uniform electric field E that has a magnitude of $8.0 \times 10^{4} \, \mathrm{V/m}$. The proton undergoes a displacement $d = 0.50 \, \mathrm{m}$ to point B in the direction of E. Calculate the speed of the proton after completing the displacement.

Solution:

In the uniform electric field proton moves with constant acceleration

$$a = \frac{F}{m} = \frac{qE}{m}.$$

The displacement is related with acceleration by equation

$$d = \frac{v^2 - v_0^2}{2a}.$$

Thus the final speed of the proton

$$v = \sqrt{v_0^2 + 2ad} = \sqrt{2ad} = \sqrt{2 \frac{qE}{m} d}.$$

Because the charge and mass of the proton are

$$q = 1.60 \times 10^{-19} \, \mathrm{C}, \quad m = 1.67 \times 10^{-27} \, \mathrm{kg}$$

we find

$$v = \sqrt{2 \times \frac{1.60 \times 10^{-19} \times 8.00 \times 10^4}{1.67 \times 10^{-27}} \times 0.50} = 2.77 \times 10^6 \, \frac{\mathrm{m}}{\mathrm{s}}.$$

Answer: $v = 2.77 \times 10^6 \, \frac{\mathrm{m}}{\mathrm{s}}$.