Question

A small charged ball is hovering in the state of equilibrium at a height H over a large horizontal uniformly charged sheet. If a disc ok radius r (r<<h) is removed from the plate directly underneath the ball, then the acceleration of the ball will be?

1. (g/2)(r²/h²)
2. (g/2)(h²/r²)
3. (g/4)(r²/h²)
4. (g/4)(h²/r²)​

Answer 1

The underneath the ball then the acceleration of the ball will be the $\frac{g}{2}(\frac{r^{2}}{h^{2}})$.

Explanation:

$E=$$\frac{O}{2\pi } [\frac{1-x}{\sqrt{r^{2}+x^{2}} } ]$

$aE=mg$

$ia=\frac{qEi}{m}$

$\frac{xq}{m}=\frac{g}{E} =\frac{g}{\alpha } (2E_{0})$

$a=(1-\frac{h}{\sqrt{r^{2}+h^{2}} } )g$

$a=[1-\frac{h}{\sqrt{1+\frac{h}{r} } } ]g$

$a=[1-\frac{h}{h\sqrt{(\frac{r}{h})^{2}+1 } }]g$

$(1+y)^{n}=1+ny$

$a=1-[(1+y^{2}^{\frac{-1}{2}}]g$

$a=1-(1-\frac{1}{2}y^{2})g$

$a=\frac{g}{2}(\frac{r^{2}}{h^{2}})$.

Answer 2

Answer:

A small charged ball is hovering in the state of equilibrium at a height H over a large horizontal uniformly charged sheet. If a disc ok radius r (r<<h) is removed from the plate directly underneath the ball, then the acceleration of the ball will be $a=\frac{g}{2}\frac{r^2}{h^2}$

Explanation:

A small charged ball remains in one place in the air at a height $H$ from a uniformly charged sheet.

If a disc of the radius $r$ is removed from the sheet directly underneath the ball.

From the question, it is given that the ball is in equilibrium, that is net force is zero.

The forces acting on the ball are gravitational force due to its mass that is acting downward and a force due to the electric field of the sheet acting upward as shown in the figure.

By equating these forces we get,

$F_{g}=F_{e}$

$mg=qE$

The electric field due to plane sheet of charge density σ  is

$E=$σ/$2$ε₀

Now, $q/m=g/E=g(2$ ε₀)/σ

If the disc is removed from the plate acceleration of the ball will be $a=F/m=qE/m$

This electric field is due to the circular disc at a distance $h$, which is equal to

$E=$σ/$2$ε₀$(1-\frac{h}{\sqrt{r^{2}+h^{2} } } )$

Take $h$ common out and cancel

$E=$σ/$2$ε₀$(1-\frac{1}{\sqrt{r^{2}/h^{2} +1} } } )$

Rearrange the equation using $(1+x)^{n}=1+nx$

$\text{.}E=\frac{\sigma }{2\varepsilon _0}\left(1-\left(\f1-\frac{r^{2} }{h^{2} } }\right)^{-\frac{1}{2}}\right).$

$E=\left(\frac{r^2}{h^2}\right)\frac{\sigma }{2\varepsilon _0}$

Then acceleration is $a=\frac{q}{m}\left(\frac{\sigma }{2\varepsilon _0}\right)\frac{r^2}{h^2}$

Substitute for $q/m$ from the first equation $a=\frac{g}{2}\frac{r^2}{h^2}$