What is the correct answer to the electric strength problem?

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The discussion centers on solving an electric strength problem related to a parallel-plate capacitor. The user initially calculated the charge using the formula q = E * ε * A but found their answer incorrect when compared to the expected unit of 1e-6c. They later realized that using a different formula, q = E * k * ε * A, provided the correct answer. The problem involves a capacitor with specific dimensions and a dielectric constant for paper. Ultimately, the user resolved their confusion and confirmed the correct approach to find the maximum charge.
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https://hw.utexas.edu/tmp/Muddam1/1077653690Xuj.pdf

I did this: I manipulated E = \frac{q}{\epsilon*A} to get q = E*\epsilon*A (E= electric strength), so 1.6e7*8.85e-12*.0246*.0501 = 1.74516e-7 but because the HW asks the answer to the 1e-6, I divided 1.74516e-7/1e-6= .174516 but it's wrong does anyone know why?
 
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Whoa...

Ok this is the problem

009 (part 1 of 2) 10 points
A parallel-plate capacitor of dimensions
2:46 cm £ 5:01 cm is separated by a 0:61 mm
thickness of paper.
Find the capacitance of this device. The
dielectric constant · for paper is 3.7. Answer
in units of pF.

I got this right.

010 (part 2 of 2) 10 points
What is the maximum charge that can be
placed on the capacitor? The electric strength
of paper is 1:6 £ 107 V=m. Answer in units of 1e-6c.
this is the one I need help on
 
...

Never mind, I got it.
 
So what's the answer?
 
I used a different formula to get the answer: q = E*k\epsilon*A
 
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