Question

A small sphere with mass $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{kg}$and charge +7.00 m Cis released from rest at a distance ofabove a large horizontal insulating sheet of charge that has uniform surface charge density is $\mathbf{\sigma }\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{pC}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. Using energy methods, calculate the speed of the sphere when it isabove the sheet.

Short answer

The speed of the sphere when it is 0.100 m above the sheet is 1.45m/s.

Step-by-step solution

Step-1: Equation for finding the speed of the sphere

The sphere is having both mass and charge, thus is affected by the gravitational force as well as electrostatic force. The total mechanical energy will be conserved at the instant of dropping it from a height of ${\mathrm{l}}_{\mathrm{a}}=0.400\mathrm{m}$, and when it reaches the height ${\mathrm{l}}_{\mathrm{b}}=0.100\mathrm{m}$above the sheet, attaining speed.

${\mathrm{U}}_{\mathrm{lb}}+{\mathrm{T}}_{\mathrm{lb}}={\mathrm{U}}_{\mathrm{lb}}+{\mathrm{T}}_{\mathrm{lb}}$

At the height ${\mathrm{l}}_{\mathrm{a}}=0.400\mathrm{m}$, the sphere is at rest, so, kinetic energy will be zero, $T_{la}=0$, but there will be gravitational potential energy and the electrostatic potential energy, which makes $U_{la}=mg{l}_a+qV$. But, the electric field is the gradient of potential, then $E=\frac{V}{d}$which gives$V=Ed$, in terms of distance$l_a$, $\mathrm{V}={\mathrm{El}}_{\mathrm{a}}$. Thus, localid="1668227458997" ${{\mathrm{E}}_{\mathrm{l}}}_{\mathrm{a}}={\mathrm{U}}_{{\mathrm{l}}_{\mathrm{a}}}={\mathrm{mgl}}_{\mathrm{a}}+{\mathrm{qEl}}_{\mathrm{a}}$, where ${\mathrm{El}}_{\mathrm{a}}$is the total mechanical energy.

At the height , the sphere has attained kinetic energy also, then the total energy has components of gravitational potential energy, electrostatic potential energy, and kinetic energy, making ${\mathrm{E}}_{\mathrm{b}}={\mathrm{U}}_{\mathrm{b}}+{\mathrm{T}}_{\mathrm{b}}=\left[{\mathrm{mgl}}_{\mathrm{b}}+\mathrm{qE}{\text{E}}_{\mathrm{b}}\right]+\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2$

But electric field due to a plane sheet is given by $\mathrm{E}=\frac{\sigma }{{\epsilon }_0}$, where is the surface charge density, and is the permittivity $={\epsilon }_0=8.854\times {10}^{-12}{\mathrm{m}}^{-3}{\mathrm{kg}}^{-1}{\mathrm{s}}^4{\mathrm{A}}^2$.

Then, ${\mathrm{E}}_{\mathrm{l}}={\mathrm{U}}_{\mathrm{b}}={\mathrm{mgl}}_{\mathrm{a}}+\mathrm{q}{\left[\frac{\sigma }{{\epsilon }_0}\right]}_{\mathrm{a}}\text{and}{\mathrm{E}}_{\mathrm{b}}={\mathrm{U}}_{\mathrm{b}}+{\mathrm{T}}_{\mathrm{b}}=\left[{\mathrm{mgl}}_{\mathrm{b}}+\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]{\mathrm{l}}_b\right]+\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2$

But as per the conservation law,

${\mathrm{E}}_{{\mathrm{l}}_{\mathrm{a}}}={\mathrm{E}}_{{\mathrm{l}}_{\mathrm{b}}}$

${\mathrm{mgla}}_{\mathrm{a}}+\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]{\mathrm{l}}_{\mathrm{a}}=\left[{\mathrm{mgl}}_{\mathrm{b}}+\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]{\mathrm{l}}_{\mathrm{b}}\right]+\frac{1}{2}{\mathrm{mv}}_{\mathrm{b}}^2$

Solve for,

$\begin{array}{r}\left.\frac{1}{2}{{\mathrm{mv}}_{\mathrm{b}}}^2={\mathrm{mgl}}_{\text{a}}+\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]{\mathrm{l}}_{\mathrm{a}}-\left({\mathrm{mgl}}_{\mathrm{b}}+\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]{\mathrm{l}}_{\mathrm{b}}\right)\right]\\ {\mathrm{v}}_{\mathrm{b}}=\sqrt{\left.\frac{2}{m}\left(\mathrm{mg}-\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]\right)\left({\mathrm{I}}_{\mathrm{a}}-{\mathrm{l}}_{\mathrm{b}}\right)\right\}}\end{array}$

Step-2: Calculation of the speed of sphere

At the final instance, ${\mathrm{l}}_{\mathrm{b}}=0.100\mathrm{m}$the speed$v_b$, is obtained by substituting the values of

$\begin{array}{r}\mathrm{m}={5.00}^{\ast }{10}^{-7}\mathrm{kg},\mathrm{q}=+7.00\mu \mathrm{C}=+7^{\ast }{10}^{-6}\mathrm{C},{\mathrm{I}}_{\mathrm{a}}=0.400\mathrm{m},{\mathrm{l}}_{\mathrm{b}}=0.100\mathrm{m}\\ \sigma =+8.00\mathrm{pC}/{\mathrm{m}}^2=+{8.00}^{\ast }{10}^{-12}\mathrm{C}/{\mathrm{m}}^2,{\epsilon }_0=8.854\times {10}^{-12}{\mathrm{m}}^{-3}{\mathrm{kg}}^{-1}{\mathrm{s}}^4{\mathrm{A}}^2,\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\\ \text{in}{\mathrm{v}}_{\mathrm{b}}=\sqrt{\left.\frac{2}{m}\left(\mathrm{mg}-\mathrm{q}\left[\frac{\sigma }{{\epsilon }_0}\right]\right)\left({\mathrm{I}}_{\mathrm{a}}-{\mathrm{I}}_{\mathrm{b}}\right)\right\}}\\ {\mathrm{v}}_{\mathrm{b}}=\sqrt{\left\{\frac{2}{5^{\ast }{10}^{-7}}\right.}\left[\left(5^{\ast }{10}^{-7\ast }9.8\right)-7^{\ast }{10}^{-6}\left[\frac{8^{\ast }{10}^{-12}}{{8.854}^{\ast }{10}^{-12}}\right](0.400-0.100)\right.\\ \left.{\mathrm{v}}_{\mathrm{b}}=\sqrt{\left\{\frac{2}{5^{\ast }{10}^{-7}}\right.}\left[\left(5^{\ast }{10}^{-7\ast }9.8\right)-\left(\frac{7^{\ast }{10}^{-6\ast }8}{8.854}\ast 0.300\right)\right]\right\}\\ {\mathrm{v}}_{\mathrm{b}}=1.45\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the sphere at a distance 0.100m from the sheet is 1.45m/s.