91Ó°ÊÓ

• The initial speed of the proton is\(9.70 \times 1{0^2}\,m/s\).
• The surface charge density of the sheet is \(2.34 \times 1{0^{ - 9}}\,C/{m^2}\).
• The proton is moving parallel to the sheet at a distance of \(0.360\,m\).

The electric field due to infinite sheet of charge is given by,

\(E = \frac{\sigma }{{2{\varepsilon _o}}}\)…… (1)

Here\(E\)is the magnitude of the electric field,\(\sigma \)is the surface charge density,\({\varepsilon _o}\)is the permittivity of free space (vacuum).

The electric force acting on the proton is given by,

\(F = eE\) ……. (2)

Here\(F\)is the magnitude of the force,\(e\)is the charge of the proton.

According to the Newton’s second law of motion, the expression for the force is given by,

\(F = ma\) …… (3)

Here\(m\)is the mass and\(a\)is the acceleration.

Now equate equation (2) and (3)

\(eE = ma\)

Now substitute\(\frac{\sigma }{{2{\varepsilon _o}}}\)for\(E\) from equation (1) into the above equation,

\(\begin{aligned}e\left( {\frac{\sigma }{{2{\varepsilon _o}}}} \right) = ma\\a = \frac{{e\sigma }}{{2{\varepsilon _o}m}}\end{aligned}\)

So, the expression for the acceleration is,

\(a = \frac{{e\sigma }}{{2{\varepsilon _o}m}}\) …… (4)

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 \(1203.3\,m/s\).