A disc of radius R is kept such that its axis coincides with the x-axis and its centre is at (d,0,0) .The thickness of disc is t and it carries a uniform volume charge density ¶ .The external electric field in the space is given by E=kr(r is position vector of any point in space with respect to the origin of the coordinate system). If the electric force on the disc is mπkdt¶R^2 /4 then find value of m?

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Solitary · Answer

Given:

radius of disc = R

thickness = t

volume charge density = $\rho$

it’s placed at (d,0,0) with its axis along the x-axis

since the thickness is t, the surface charge density of each surface can be given as
$\sigma = \rho t$

now on the surface, consider a small ring element dr at a distance r from the centre

the area covered by this ring can be given by $2\pi rdr$
thus the charge

$dq = \sigma\cdot2\pi r dr = \rho t \cdot 2\pi r dr$

now, the force can be given as
$dF = E\cdot dq$
$dF = Kd \cdot 2\pi \rho t rdr$
integrating both sides,
$\int_0^F dF = \int_0^R Kd \cdot 2\pi \rho t rdr$

$F = 2 \pi Kdt\rho [\frac{r^2}{2}]_0^R$

$F = \pi K dt \rho R^2$
comparing this with $F = m\pi K dt \rho R^2$

m = 1

Hence, m=1 is the required solution.

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