Question

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \times 10^{ - 8} {\rm{ C/m}}^2. The object is initially 0.570 m from the sheet.
What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.260 m?

Answer 1

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$