Answer :
Answer:
The strength of the field is 22.84 N/C.
Explanation:
Given that,
Speed [tex]v= 7.5\times10^{5}\ m/s[/tex]
Distance = 7.0 cm
We need to calculate the acceleration
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Put the value in the equation
[tex]0-(7.5\times10^{5})^2=2\times a\times7.0\times10^{-2}[/tex]
[tex]a =-\dfrac{(7.5\times10^{5})^2}{2\times7.0\times10^{-2}}[/tex]
[tex]a =-4.017\times10^{12}\ m/s^2[/tex]
We need to calculate the strength of the field
Using newton's second law and electric force
[tex]F = ma = qE[/tex]
[tex]-qE=-ma[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
Put the value into the formula
[tex]E=\dfrac{9.1\times10^{-31}\times4.017\times10^{12}}{1.6\times10^{-19}}[/tex]
[tex]E=22.84\ N/C[/tex]
Hence, The strength of the field is 22.84 N/C.