## 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

1. Michele_Laino

we have to compute this quantity: $\Large E\left( r \right) = K\frac{Q}{{{r^2}}}$ for all of your four cases

2. anonymous

ok! how can we do that?

3. Michele_Laino

for example, first case: $E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 1}}} \right)}^2}}} = \frac{{270}}{{16}}$

4. anonymous

and we get 16.875?

5. Michele_Laino

oops.. I have made an error: $E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{27 \times {{10}^9}}}{{16}}$

6. anonymous

ok! so we get 1687500000?

7. Michele_Laino

second case: $E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{6 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}$

8. anonymous

3375000000?

9. Michele_Laino

third case: $E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{3 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{108 \times {{10}^9}}}{{16}}$

10. anonymous

and then 6750000000 ?

11. Michele_Laino

fourth case: $E\left( r \right) = K\frac{Q}{{{r^2}}} = 9 \times {10^9}\frac{{1.5 \times {{10}^{ - 6}}}}{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}} = \frac{{54 \times {{10}^9}}}{{16}}$

12. Michele_Laino

what is the highest value?

13. anonymous

the highest value is case 3? with 108*10^9 / 16 ?

14. Michele_Laino

yes! that's right!

15. anonymous

yAY! what does that mean about the solution? :/ oi am not sure which choice it would be :/

16. Michele_Laino

it is the third option as you have said before

17. anonymous

ohh okay! i see:) thank you!! onto the next one then :)

18. Michele_Laino

ok!