Question

22.32: A very small object with mass \(8.20 \times 1{0^{ - 9}}\,kg\) and positive charge \(6.50 \times 1{0^{ - 9}}\,C\) is projected directly towards a very large insulating sheet of positive charge that has uniform surface charge density \(5.90 \times 1{0^{ - 8}}\,C/{m^2}\). The object is initially \(0.400\,m\) from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be \(0.100\,m\)?

Short answer

The initial speed of the object is \(39.8\,m/s\).

Step-by-step solution

Identification of given data

• The mass of the object is\(8.20 \times 1{0^{ - 9}}\,kg\).
• The magnitude of the charge of the object is\(6.50 \times 1{0^{ - 9}}\,C\).
• The surface charge density of the sheet is \(5.90 \times 1{0^{ - 8}}\,C/{m^2}\).
• Initial distance of the object from the sheet is \(0.400\,m\).

Understanding the concept:

The expression for the acceleration in terms of surface charge density, charge and mass is given by,

\(a = \frac{{q\sigma }}{{2{\varepsilon _o}m}}\)…… (1)

Here \(q\) is the magnitude of the charge, \(\sigma \) is the surface charge density, \(m\) is the mass of the object, \({\varepsilon _o}\) is the permittivity of free space.

Determine the initial speed of the object:

Using the kinematics, equation of motion is given by,

\({v^2} = {u^2} - 2as\)

Overall displacement of the object,

\(\begin{aligned}{c}s = 0.4 - 0.1\\ = 0.3\,m\end{aligned}\)

Since the object is trying to approach the large insulating sheet, so the final speed of the object is zero,

\(\begin{aligned}0 = {u^2} - 2\frac{{q\sigma }}{{2{\varepsilon _o}m}}s\\{u^2} = \frac{{q\sigma }}{{{\varepsilon _o}m}}s\end{aligned}\)

Substitute\(6.50 \times 1{0^{ - 9}}\,C\)for\(q\), \(5.90 \times 1{0^{ - 8}}\,C/{m^2}\)for\(\sigma \),\(0.3\,m\)for\(s\),\(8.85 \times 1{0^{ - 12}}\,F/m\)for\({\varepsilon _o}\) and\(8.20 \times 1{0^{ - 9}}\,kg\)for\(m\)into the above equation,

\(\begin{aligned}{u^2} = \frac{{\left( {6.50 \times 1{0^{ - 9}}\,C} \right)\left( {5.90 \times 1{0^{ - 8}}\,C/{m^2}} \right)}}{{\left( {8.85 \times 1{0^{ - 12}}\,F/m} \right)\left( {8.20 \times 1{0^{ - 9}}\,kg} \right)}}\left( {0.3\,m} \right)\\ = 1585.36\end{aligned}\)

\(\therefore u = 39.8\,m/s\)

Therefore the initial speed of the object is \(39.8\,m/s\).