Chapter 16, Problem 63GP

To determine

To Write: An expression for electric field strength as function of time.

Answer to Problem 63GP

  $\frac{1.08\times {10}^7}{{[3-\cos \left(12.54\right)t]}^2}\text{N}/\text{C}$

Explanation of Solution

Given:

The spring constant of an ideal spring, $k=126\text{N}/\text{m}$ .

Mass of the sphere, $m=0.800\text{kg}$ .

Pulled distance from the equilibrium gives amplitude, $A=0.05\text{m}$ .

Electric charge, $Q=-3\times {10}^{-6}\text{C}$ .

Formula used:

Electric field strength due to charge Q at a distance r away from it can be expressed as follows:

  $E=\frac{1}{4\pi {ϵ}_0}\cdot \frac{\underset{\_}{Q}}{r^2}$

Here, ${ϵ}_0$ is the absolute permittivity.

Calculation:

The sphere pulled and so it undergoes simple harmonic motion.

The equation of simple harmonic motion can be written as

  $x=A\cos \omega t$

Here, A is the amplitude, $\omega$ is the angular frequency and t is the time.

  $\begin{array}{l}\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{126\text{N}/\text{m}}{0.800\text{kg}}}\\ \omega =12.54\text{rad/s}\end{array}$

So,

  $\begin{array}{l}x=A\cos \omega t\\ x=0.05\cos \left(12.54t\right)\end{array}$

The charge given to the sphere is negative and thus, the electric field at the table will be directed upwards.

Initial distance of sphere’s center from the table top $=15.0\text{cm}=0.15\text{m}$

Distance of the sphere as a function of time would be:

  $r=(0.15\text{m})-(0.05\text{m})[\cos (12.54)t]$

Thus, the electric field would be:

  $E=\frac{1}{4\pi {ϵ}_0}\cdot \frac{\underset{\_}{Q}}{r^2}$

  $\begin{array}{l}=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(3\times {10}^{-6}\text{C}\right)}{{\left\{(0.15\text{m})-(0.05\text{m})\left[\cos \left(12.54{\text{s}}^{-1}\right)t\right]\right\}}^2}\hfill \\ =\frac{27\times {10}^3}{{(0.05)}^2{[3-\cos (12.54)t]}^2}\hfill \\ =\frac{1.08\times {10}^7}{{[3-\cos (12.54)t]}^2}\text{N}/\text{C}\hfill \end{array}$

The direction of electric field is upward.