Question

Electric flux question

Answer 1

I calculated the total flux as -780.33 then multiplied it by epsilon naught.

Answer 2

here we have to apply this formula:
\[\Phi \left( {\mathbf{E}} \right) = \frac{q}{{{\varepsilon _0}}}\]

Answer 3

The two different electric fields are what confuse me.

Answer 4

Now, the requested flux at the left side, is:

\[\Phi \left( {\mathbf{E}} \right) = - \left( {36 + 21} \right) \cdot {3.7^2} = - 780.33\frac{{N{m^2}}}{C}\]

Answer 5

the situation is this:

Answer 6

as we can see we have two negative terms

Answer 7

I understand

Answer 8

now, the requested charge is:
\[q = {\varepsilon _0} \cdot \Phi \left( {\mathbf{E}} \right) = 8.85 \cdot {10^{ - 12}} \cdot \left( { - 780.33} \right) = ...?\]

Answer 9

-6.91e-9 C

Answer 10

yes!

Answer 11

I think it is the right result

Answer 12

thank you so much! really appreciate it

Answer 13

:)