Answer:
[tex]E=101955.8volt/m[/tex]
The electric field direction is toward the right
Explanation:
From the question we are told that
Initial X-co-ordinate of proton [tex]X_1=20.0cm => \frac{20}{100}m[/tex]
Initial speed of Proton [tex]V_1= 3.5*10^6 m/s[/tex]
Final X-co-ordinate of proton [tex]X_2=80.0cm => \frac{80.0}{100}m[/tex]
Final speed of Proton [tex]V_2=0[/tex]
Generally the mass of Proton is given by
[tex]m_P=1.67*10^-27[/tex]
Generally the kinetic energy of the proton is mathematically given by
[tex]K.E_p=1/2mv^2[/tex]
[tex]K.E_p=1/2*1.6*10^-^2^7*(3.5*10^6)^2[/tex]
[tex]K.E_p=9.8*10^-^1^5[/tex]
Generally the change in electric potential [tex]\triangle V[/tex] is mathematically given by
[tex]\triangle V =\frac{K.E_p}{q}[/tex]
Charge on a proton [tex]q=1.602*10^-^1^9[/tex]
[tex]\triangle V =\frac{9.8*10^-^1^5}{1.602*10^-^1^9}[/tex]
[tex]\triangle V =61173.5volts[/tex]
Generally the equation for magnitude of an electric field is mathematically given by
[tex]E=\frac{\triangle V}{\triangle d}[/tex]
Where
[tex]d=0.8m-0.2m\\d=0.6m[/tex]
Therefore
[tex]E=\frac{61173.5}{0.8-0.2}[/tex]
[tex]E=\frac{61173.5}{0.6}[/tex]
[tex]E=101955.8volt/m[/tex]
The direction of the charge is towards the right
