Question

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

Answer 1

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

Answer 2

ok! how can we do that?

Answer 3

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

Answer 4

and we get 16.875?

Answer 5

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

Answer 6

ok! so we get 1687500000?

Answer 7

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

Answer 8

3375000000?

Answer 9

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

Answer 10

and then 6750000000 ?

Answer 11

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

Answer 12

what is the highest value?

Answer 13

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

Answer 14

yes! that's right!

Answer 15

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

Answer 16

it is the third option as you have said before

Answer 17

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

Answer 18

ok!