Chapter 16, Problem 31P

To determine

To find: The direction and magnitude of field.

Answer to Problem 31P

  $E=6.54\times {10}^{-10}\text{N}/\text{C}$ acting towards south.

Explanation of Solution

Given:

Acceleration towards north $a=115{\text{m/s}}^{\text{2}}$

Formula used:

The acceleration is defined as,

  $a=\frac{eE}{m_e}$

Calculation:

Mass of an electron $m_e=9.11\times {10}^{-31}\text{kg}$

Charge on an electron $e=-1.6\times {10}^{-19}\text{C}$

Acceleration $a=115\text{m}/{\text{s}}^2$ towards north

Let the applied electric field = E

Then $a=\frac{eE}{m_e}$

  $\begin{array}{l}E=\frac{a{m}_e}{e}\\ =\frac{115\text{m}/{\text{s}}^2\times 9.11\times {10}^{-31}}{-1.6\times {10}^{-19}}\\ =6.54\times {10}^{-10}\text{N}/\text{C}\end{array}$

Or,

  $E=6.54\times {10}^{-10}\text{N}/\text{C}$ acting towards south, as the acceleration of the electron is in the opposite direction of the electric field.

Conclusion:

Therefore, $E=6.54\times {10}^{-10}\text{N}/\text{C}$ acting towards south.