Question

Find the force on a point charge of \(30 \; \mu C\) at \((0,0,5) \;m\) due to a \(4 \; m\) square in the \(z=0\) plane between \(x = \pm 2\;m\) and \(y = \pm 2\;m\) with a total charge of \(500 \; \mu C\), distributed uniformly..

Answer 1

As, \(500 \mu C\) is distributed on the square uniformly, so we can find its Surface Charge Density as:

\[\rho_s = \frac{Q}{Area} = \frac{500 \times 10^{-6}}{(4)^2} = 31.25\; \; \mu C/m^2\]

Answer 2

Then I consider a small area ds on that square and take the point on that surface as (x,y,0) and then finding the distance between (0,0,5) and (x,y,0) we get:

\[r = -xa_x -ya_y + 5a_z\]

Answer 3

As ds = dx.dy so finally we get the expression for small force dF as:

\[\large dF = \frac{\rho_{s} \cdot Q_2 \cdot dx \cdot dy[-xa_x -ya_y +5a_z]}{4 \pi \epsilon_o[(x^2 + y^2 + 25)^{3/2}]}\]

Answer 4

Now the only work left is to integrate this typical integral in the limits of x and y as -2 to +2.

\[\large F = \int\limits_{-2}^{2} \int\limits_{-2}^{2}dF\]

Answer 5

If I am not wrong then, integral with \(-xa_x\) and with \(-ya_y\) must be 0, and the remaining term left finally must be according to me is:

\[\large \color{blue}{F = \frac{5 \cdot \rho_s \cdot Q_2 \cdot a_z}{4 \pi \epsilon_o} \int\limits_{-2}^{2} \int\limits_{-2}^{2} \frac{dx \cdot dy}{(x^2 + y^2 + 25)^{3/2}}}\]

Answer 6

@eliassaab

Answer 7

@experimentX

Answer 8

@experimentX, do some of your Experiments here at this post also... :)

Answer 9

After solving it more, if evaluate firstly for dx integrand, then:

Put \(x = \sqrt{y^2 + 25} \cdot tan(\theta)\)

And proceeding this way and plugging in the limits for x, I am finally getting:

\[\large \color{red}{F = \frac{5 \cdot \rho_s \cdot Q_2 \cdot a_z}{4 \pi \epsilon_o} \int\limits_{-2}^{2}\frac{4 \cdot dy}{(y^2 + 25) \cdot \sqrt{y^2 + 29}}}\]

Answer 10

Note that \(a_x, a_y, a_z\) all are unit vectors along x, y and z directions..

@hartnn, how to integrate this integral now??

Answer 11

@waterineyes, I have little to no background in physics except what mechanics I've learned in calculus, so I'll assume all your work up to this point is accurate.

I'll also ignore the constant because it can always be factored in later. For the integral, you can use a trig substitution.
\[y=\sqrt{29}\tan u~~\Rightarrow~~dy=\sqrt{29}\sec^2u~du\]
Then you have
\[\int\limits_{a}^{b}\frac{\sqrt{29}\sec^2u}{\left(\left(\sqrt{29}\tan u\right)^2 + 25\right) \sqrt{\left(\sqrt{29}\tan u\right)^2 + 29}}~du\]
where \(a=\tan^{-1}\left(-\dfrac{2}{\sqrt{29}}\right)\) and \(b=\tan^{-1}\left(\dfrac{2}{\sqrt{29}}\right)\).
\[\int\limits_{a}^{b}\frac{\sqrt{29}\sec^2u}{\left(29\tan^2u + 25\right) \sqrt{29\tan^2u + 29}}~du\\
\int\limits_{a}^{b}\frac{\sec^2u}{\left(29\tan^2u + 25\right) \sqrt{\tan^2u + 1}}~du\\
\int\limits_{a}^{b}\frac{\sec^2u}{\left(29\tan^2u + 25\right) \sqrt{\sec^2u}}~du\\
\int\limits_{a}^{b}\frac{\sec u}{29\tan^2u + 25}~du\\
\int\limits_{a}^{b}\frac{du}{29\dfrac{\sin^2u}{\cos u} + 25\cos u}\\
\int\limits_{a}^{b}\frac{\cos u}{29\sin^2u + 25\cos^2 u}~du\\
\int\limits_{a}^{b}\frac{\cos u}{4\sin^2u+1}~du\]
Then substitute \(t=\sin u\) so that \(dt=\cos u~du\):
\[\int\limits_{\alpha}^{\beta}\frac{dt}{4t^2+1}\]
with \(\alpha=\sin a=-\dfrac{2}{\sqrt{33}}\) and \(\beta=\sin b=\dfrac{2}{\sqrt{33}}\).

Another trig sub should do it - let \(t=\dfrac{1}{2}\tan x\) so \(dt=\dfrac{1}{2}\sec^2x~dx\):
\[\int\limits_{\gamma}^{\delta}\frac{\dfrac{1}{2}\sec^2x}{4\left(\dfrac{1}{2}\tan x\right)^2+1}~dx\]
where \(\gamma=\tan^{-1}2\alpha\) and \(\delta=\tan^{-1}2\beta\).
\[\frac{1}{2}\int\limits_{\gamma}^{\delta}\frac{\sec^2x}{\tan^2x+1}~dx\\
\frac{1}{2}\int\limits_{\gamma}^{\delta}dx\]

Answer 12

Chances are I made some slight error along the way, but the method remains the same.

Answer 13

There is indeed a mistake. For the last two lines of the [a,b] integral w.r.t. \(u\), the denominator should be
\[4\sin^2u+\color{red}{25}\]
which means you would use
\[t=\dfrac{5}{2}\tan x\]

Answer 14

I calculated the expression as
\[ \frac{Q}{\pi \epsilon_0} \int_{-\pi/4}^{\pi/4} \frac{\cos \left( \arctan \left( 2\sec \theta/5\right)\right)}{\sqrt{4 \sec^2 \theta + 25}} \sec (\theta) d\theta \]