Five charges are placed in a closed box. Each
charge (except the ﬁrst) has a magnitude
which is twice that of the previous one placed
in the box. All charges have the same sign
and (after all the charges have been placed in
the box) the net electric ﬂux through the box
is 9.5 × 10
7
N · m2
/C.
What is the magnitude of the smallest
charge in the box? The permittivity of a
vacuum is 8.85419 × 10
−12
C
2
/N · m2
.
Answer in units of µC

Question

Five charges are placed in a closed box. Each
charge (except the ﬁrst) has a magnitude
which is twice that of the previous one placed
in the box. All charges have the same sign
and (after all the charges have been placed in
the box) the net electric ﬂux through the box
is 9.5 × 10
7
N · m2
/C.
What is the magnitude of the smallest
charge in the box? The permittivity of a
vacuum is 8.85419 × 10
−12
C
2
/N · m2
.
Answer in units of µC

jamesj · Answer

Remember the Gaussian result that the integral of the flux must be equal to the sum of the charges
$$ \int E \cdot dA = \frac{\sum Q}{\epsilon_0} $$

Now you know the left hand side.  Use that to solve for the RHS.

anonymous · Answer

^^ Do what JamesJ said

turingtest · Answer

yep!

jamesj · Answer

Notice that the sum of the charges is 
$$ q + 2q + 4q + 8q + 16q = 31q $$

anonymous · Answer

so (9.5 * 10^7)(8.85 *10^-12)/31

jamesj · Answer

Looks about right.

anonymous · Answer

my answer is 2,713 * 10^-5. imentering this in the system but its coming wrong any ideas why?

jamesj · Answer

Given the data you provided, that is the answer.

turingtest · Answer

^agreed