Question

Surface charge density data, $\rho_s(r)$, in $[C/m^2]$, versus radius, $r$ in $[m]$, on the upper patch: $\begin{array}{|c|c|} \hline r (m) & \rho_s(r), [C/m^2], upper \\ \hline 0.00E+00 & 1.075E-07 \\ 1.00E-05 & 1.085E-07 \\ 2.00E-05 & 1.105E-07 \\ 3.00E-05 & 1.135E-07 \\ 4.00E-05 & 1.175E-07 \\ 5.00E-05 & 1.235E-07 \\ 6.00E-05 & 1.336E-07 \\ 7.00E-05 & 1.486E-07 \\ 8.00E-05 & 1.767E-07 \\ 9.00E-05 & 2.420E-07 \\ 1.00E-04 & 1.085E-06 \\ \hline \end{array}$ 

          Surface charge density data, $\rho_s(r)$, in
$[C/m^2]$, versus radius, $r$ in $[m]$, on the
upper patch:

$\begin{array}{|c|c|}
\hline
r (m) & \rho_s(r), [C/m^2], upper \\
\hline
0.00E+00 & 1.075E-07 \\
1.00E-05 & 1.085E-07 \\
2.00E-05 & 1.105E-07 \\
3.00E-05 & 1.135E-07 \\
4.00E-05 & 1.175E-07 \\
5.00E-05 & 1.235E-07 \\
6.00E-05 & 1.336E-07 \\
7.00E-05 & 1.486E-07 \\
8.00E-05 & 1.767E-07 \\
9.00E-05 & 2.420E-07 \\
1.00E-04 & 1.085E-06 \\
\hline
\end{array}$
Show more…
Surface charge density data, (r), in [C/m^2], versus radius, r in [m], on the upper patch: r (m) (r), [C/m^2], upper 0.00E+00 1.075E-07 1.00E-05 1.085E-07 2.00E-05 1.105E-07 3.00E-05 1.135E-07 4.00E-05 1.175E-07 5.00E-05 1.235E-07 6.00E-05 1.336E-07 7.00E-05 1.486E-07 8.00E-05 1.767E-07 9.00E-05 2.420E-07 1.00E-04 1.085E-06

Added by Nicole J.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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I'm supposed to calculate the total charge of the disc using the Trapezoidal Rule modified for use with Cartesian coordinates. The equation I'm using for each ring is π(change in R)(charge of inner radius + charge of outer radius). The total charge is supposed to be 7.739 x 10^-15, but I am getting 7.19 x 10^-15. What is wrong with my equation? Surface charge density data, Os(r), in [C/m], versus radius, r in [m], on the upper patch: r (m) pr) [C/m], upper 1.075E-07 1.085E-07 0.00E+00 1.00E-05 2.00E-05 1.105E-07 3.00E-05 1.135E-07 4.00E-05 1.175E-07 5.00E-05 1.235E-07 6.00E-05 1.336E-07 7.00E-05 1.486E-07 8.00E-05 1.767E-07 9.00E-05 2.420E-07 1.00E-04 1.085E-06
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Transcript

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00:01 So now consider a gaussian surface with radius small r and length small l inside this inside the cylindrical volume of radius capital r.
00:51 And then the q enclosed will become equal to row into v.
00:57 So where row is this and the v is the volume so it will become pi r square l right so since this is the key in closed now we can write flux is equal to closed integral eda so the area will become equal to 2 pi r l now according to law we can write phi is equal to q enclosed over epsilon not let us substitute the values here then e into 2 pi r l will become equal to row into pi r square l over epsilon not now if we simplify this we will get the value of e as e will become equal to row over 2 epsilon not into r okay so we are getting this value row r over 2 epsilon not we know this is for inside so r doth and all equal to r let us solve for second part now if we solve for second part we can write we need to consider a gaussian surface.
02:36 Now to find electric field outside, to find electric field outside the cylindrical volume.
02:59 What we should do we need to consider a gaussian surface of radius small r and length small l outside the cylindrical volume.
03:26 So now the q enclosed will become equal to row into pi r square into l...
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