Respuesta :
Answer:
[tex]v=1.6\times10^{5}m/s[/tex]
Explanation:
The equation that relates electric field strength and force that a charge experiments by it is F=qE.
Newton's 2nd Law states F=ma, which for our case will mean:
[tex]a=\frac{F}{m}=\frac{qE}{m}[/tex]
We know from accelerated motion that [tex]v^2=v_0^2+2ad[/tex]
In our case the proton is released from rest, so [tex]v_0=0m/s[/tex] and we get [tex]v=\sqrt{2ad}[/tex]
Substituting, we get our final velocity equation:
[tex]v=\sqrt{\frac{2qEd}{m}}[/tex]
For the proton we know that [tex]q=1.6\times10^{-19}C[/tex] and [tex]m=1.67\times10^{-27}kg[/tex]. Writing in S.I. d=0.0025m, we obtain:
[tex]v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s[/tex]
The speed is defined as the rate of change of circular distance traveled. The proton's speed when it reaches the negative plate will be 1.6 ×10⁵ m/sec.
What is electric field strength?
The electric field strength is defined as the ratio of electric force and charge. Mathematically it is guven by;
F=qE
The given data in the problem is;
E is the electric field strength = 5.50×10^4 N/C
d is the spacing between the plates=2.50 mm
v is the speed of proton=?
From Newton's second law;
[tex]\rm a= \frac{F}{m}=\frac{qE}{m}[/tex]
If the acceleration is constant. According to the second equation of motion;
[tex]\rm v^2 = v_0^2 + 2ad \\\\\ \rm v=\sqrt{2ad}[/tex]
The value of the velocity is found by the formula;
[tex]\rm v= \sqrt{\frac{2qEd}{m} } \\\\ \rm v= \sqrt{\frac{2\times 1.6 \times 10^{-19}(5.5\times 10^4|times(0.0025)}{1.67 \times 10^{-27} }[/tex]
[tex]\rm v= 162318.5 = 1.6 \times 10^5 \ m/sec.[/tex]
Hence the proton's speed when it reaches the negative plate will be 1.6 ×10⁵ m/sec.
To learn more about the electric field strength refer to the link;
https://brainly.com/question/4264413
