Question 159: Electrostats & Capacitance

Question

Two large vertical and parallel metal plates having a separation of $$1$$ $$cm$$ are connected to a $$DC$$ voltage source of potential difference $$X$$. A proton is released at rest midway between the two plates. It is found to move at $${45^ \circ }$$ to the vertical JUST after release. Then $$X$$ is nearly

Accepted Answer

Given that the proton moves at a $45^{\circ}$ angle to the vertical, we can derive the electric field (E) as follows :
$ qE = mg $ or $ E = \frac{mg}{q} $
Since $ E = \frac{X}{d} $, we can substitute to find $ X $ :
$ \frac{X}{d} = \frac{mg}{q} $ or $ X = \frac{mgd}{q} $
Where :
$ m = 1.67 	imes 10^{-27} \, \mathrm{kg} $
$ g = 10 \, \mathrm{m/s^2} $
$ d = 1 \, \mathrm{cm} = 1 	imes 10^{-2} \, \mathrm{m} $
$ q = 1.6 	imes 10^{-19} \, \mathrm{C} $
Substituting these values, we get :
$ \begin{aligned}    X &= \frac{1.67 	imes 10^{-27} 	imes 10 	imes 1 	imes 10^{-2}}{1.6 	imes 10^{-19}} \, \mathrm{V} \\    &= \frac{1.67}{1.6} 	imes 10^{-9} \, \mathrm{V} \\    &= 1 	imes 10^{-9} \, \mathrm{V}  \end{aligned} $

Answer

$$1 	imes {10^{ - 5}}\,\,V$$

Answer

$$1 	imes {10^{ - 7}}\,\,V$$

Answer

$$1 	imes {10^{ - 9}}\,\,V$$

Answer

$$1 	imes {10^{ - 10}}\,\,V$$

Electrostats & Capacitance

Two large vertical and parallel metal plates having a separation of $1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at {45^ \circ } to the vertical JUST after release. Then X$ is nearly

Choose the correct answer: