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anonymous

  • one year ago

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

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  1. Michele_Laino
    • one year ago
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    we have to compute this quantity: \[\Large E\left( r \right) = K\frac{Q}{{{r^2}}}\] for all of your four cases

  2. anonymous
    • one year ago
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    ok! how can we do that?

  3. Michele_Laino
    • one year ago
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    for example, first case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 1}}} \right)}^2}}} = \frac{{270}}{{16}}\]

  4. anonymous
    • one year ago
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    and we get 16.875?

  5. Michele_Laino
    • one year ago
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    oops.. I have made an error: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{27 \times {{10}^9}}}{{16}}\]

  6. anonymous
    • one year ago
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    ok! so we get 1687500000?

  7. Michele_Laino
    • one year ago
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    second case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{6 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}\]

  8. anonymous
    • one year ago
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    3375000000?

  9. Michele_Laino
    • one year ago
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    third case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{108 \times {{10}^9}}}{{16}}\]

  10. anonymous
    • one year ago
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    and then 6750000000 ?

  11. Michele_Laino
    • one year ago
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    fourth case: \[E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{1.5 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}\]

  12. Michele_Laino
    • one year ago
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    what is the highest value?

  13. anonymous
    • one year ago
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    the highest value is case 3? with 108*10^9 / 16 ?

  14. Michele_Laino
    • one year ago
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    yes! that's right!

  15. anonymous
    • one year ago
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    yAY! what does that mean about the solution? :/ oi am not sure which choice it would be :/

  16. Michele_Laino
    • one year ago
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    it is the third option as you have said before

  17. anonymous
    • one year ago
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    ohh okay! i see:) thank you!! onto the next one then :)

  18. Michele_Laino
    • one year ago
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    ok!

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