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Question: A proton is moving parallel to a uniform electric field. The electric field accelerates the proton a...

A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear momentum to 5.0 × 10−23^{-23} kg. m/s from 1.5 × 10−23^{-23} kg.m/s in a time of 7.0 × 10−6^{-6} s. What is the magnitude of the electric field?

A

31.25 N/C

B

40 N/C

C

22.5 N/C

D

108 N/C

Answer

31.25 N/C

Explanation

Solution

The initial linear momentum of the proton is pi=1.5×10−23p_i = 1.5 \times 10^{-23} kg.m/s.
The final linear momentum of the proton is pf=5.0×10−23p_f = 5.0 \times 10^{-23} kg.m/s.
The time taken for this change is Δt=7.0×10−6\Delta t = 7.0 \times 10^{-6} s.

The change in linear momentum is Δp=pf−pi\Delta p = p_f - p_i.
Δp=(5.0×10−23)−(1.5×10−23)=(5.0−1.5)×10−23=3.5×10−23\Delta p = (5.0 \times 10^{-23}) - (1.5 \times 10^{-23}) = (5.0 - 1.5) \times 10^{-23} = 3.5 \times 10^{-23} kg.m/s.

According to Newton's second law, the force acting on the proton is equal to the rate of change of its linear momentum:
F=ΔpΔtF = \frac{\Delta p}{\Delta t}
F=3.5×10−23 kg.m/s7.0×10−6 sF = \frac{3.5 \times 10^{-23} \text{ kg.m/s}}{7.0 \times 10^{-6} \text{ s}}
F=3.57.0×10−23−(−6)=0.5×10−17=5.0×10−18F = \frac{3.5}{7.0} \times 10^{-23 - (-6)} = 0.5 \times 10^{-17} = 5.0 \times 10^{-18} N.

The force on a charged particle with charge qq in a uniform electric field EE is given by F=qEF = qE.
The charge of a proton is q=+e≈1.6×10−19q = +e \approx 1.6 \times 10^{-19} C.
We can find the magnitude of the electric field EE using the formula E=FqE = \frac{F}{q}.
E=5.0×10−18 N1.6×10−19 CE = \frac{5.0 \times 10^{-18} \text{ N}}{1.6 \times 10^{-19} \text{ C}}
E=5.01.6×10−18−(−19)=5.01.6×101E = \frac{5.0}{1.6} \times 10^{-18 - (-19)} = \frac{5.0}{1.6} \times 10^{1}
E=501.6=50016=1254=31.25E = \frac{50}{1.6} = \frac{500}{16} = \frac{125}{4} = 31.25 N/C.