anonymous
  • anonymous
Which of the following locations has the largest electric field? A, 0.4 cm from a +3.0 μc point charge B. 0.4 cm from a +6.0 μc point charge C. 0.2 cm from a +3.0 μc point charge D. 0.2 cm from a +1.5 μc point charge
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
we have to compute this quantity: \[\Large E\left( r \right) = K\frac{Q}{{{r^2}}}\] for all of your four cases
anonymous
  • anonymous
ok! how can we do that?
Michele_Laino
  • Michele_Laino
for example, first case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 1}}} \right)}^2}}} = \frac{{270}}{{16}}\]

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anonymous
  • anonymous
and we get 16.875?
Michele_Laino
  • Michele_Laino
oops.. I have made an error: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{27 \times {{10}^9}}}{{16}}\]
anonymous
  • anonymous
ok! so we get 1687500000?
Michele_Laino
  • Michele_Laino
second case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{6 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}\]
anonymous
  • anonymous
3375000000?
Michele_Laino
  • Michele_Laino
third case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{108 \times {{10}^9}}}{{16}}\]
anonymous
  • anonymous
and then 6750000000 ?
Michele_Laino
  • Michele_Laino
fourth case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{1.5 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}\]
Michele_Laino
  • Michele_Laino
what is the highest value?
anonymous
  • anonymous
the highest value is case 3? with 108*10^9 / 16 ?
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
yAY! what does that mean about the solution? :/ oi am not sure which choice it would be :/
Michele_Laino
  • Michele_Laino
it is the third option as you have said before
anonymous
  • anonymous
ohh okay! i see:) thank you!! onto the next one then :)
Michele_Laino
  • Michele_Laino
ok!

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