Using the kinematic equation in perpendicular direction of the insulating sheet, the expression for the speed of the proton is given by,
\(v = u + at\)
Since the proton is moving parallel to the sheet initially, so the initial speed of the proton perpendicular to the sheet is equal to zero.
Substitute\(\frac{{e\sigma }}{{2{\varepsilon _o}m}}\)for\(a\),\(0\,m/s\)for\(a\)into the above equation.
\(\begin{aligned}v = 0 + \frac{{e\sigma }}{{2{\varepsilon _o}m}}t\\ = \frac{{e\sigma }}{{2{\varepsilon _o}m}}t\end{aligned}\)
Substitute\(8.85 \times 1{0^{ - 12}}\,F/m\)for\({\varepsilon _o}\),\(1.67 \times 1{0^{ - 27}}\,kg\)for\(m\),\(2.34 \times 1{0^{ - 9}}\,C/{m^2}\)for\(t\),\(1.6 \times 1{0^{ - 19}}\,C\)for\(e\),\(2.34 \times 1{0^{ - 9}}\,C/{m^2}\) for\(\sigma \) into the above equation.
\(\begin{aligned}v = \frac{{\left( {1.6 \times 1{0^{ - 19}}\,C} \right)\left( {2.34 \times 1{0^{ - 9}}\,C/{m^2}} \right)}}{{2\left( {1.67 \times 1{0^{ - 27}}\,kg} \right)\left( {8.85 \times 1{0^{ - 12}}\,F/m} \right)}}\left( {2.34 \times 1{0^{ - 9}}\,C/{m^2}} \right)\\ = 683.974\,m/s\end{aligned}\)
Therefore the overall resultant speed of the proton is,
\(\begin{aligned}{v_R} = \sqrt {{{\left( {990} \right)}^2} + {{\left( {683.974} \right)}^2}} \\ = 1203.3\,m/s\end{aligned}\)
Therefore the speed of proton is \(1203.3\,m/s\).
