Question

(II) An electron starts from rest 24.5 cm from a fixed point charge with \({\bf{Q = - 6}}{\bf{.50}}\;{\bf{nC}}\). How fast will the electron be moving when it is very far away?

Short answer

The velocity of the electron is \(9.16 \times {10^6}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

Step-by-step solution

Step 1:Understanding of potential energy

The value of the potential energy between two point charges can be calculated by examining the value of the magnitude of both charges and the separation between the charges.

Given information

The magnitude of the charge is,\({Q_1} = - 6.50\;{\rm{nC}}\).

The charge on the electron is, \({Q_2} = - 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

The distance between the point charge and electron is, \(r = 24.5\;{\rm{cm}}\).

Evaluation of the potential energy

The potential energy between the electron and the point charge can be calculated as:

\(PE = \frac{{k{Q_1}{Q_2}}}{r}\)

Here, k is the Coulomb’s constant.

Substitute the values in the above expression.

\(\begin{aligned}PE &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( { - {\rm{6}}{\rm{.50}}\;{\rm{nC}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{nC}}}}} \right)\left( { - 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{\left( {{\rm{24}}{\rm{.5}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}\\ &= 3.82 \times {10^{ - 17}}\;{\rm{J}}\end{aligned}\)

Evaluation of the velocity of the electron

Now apply the conservation of energy principle to calculate the velocity of the electron.

\(\begin{aligned}KE &= PE\\\frac{1}{2}m{v^2} &= PE\\v &= \sqrt {\frac{{2PE}}{m}} \end{aligned}\)

Here, \(m\) is the mass of the electron and its value is \(9.11 \times {10^{ - 31}}\;{\rm{kg}}\).

Substitute the values in the above equation.

\(\begin{aligned}v &= \sqrt {\frac{{2\left( {3.82 \times {{10}^{ - 17}}\;{\rm{J}}} \right)}}{{\left( {9.11 \times {{10}^{ - 31}}\;{\rm{kg}}} \right)}}} \\ &= 9.16 \times {10^6}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the velocity of the electron is \(9.16 \times {10^6}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).