Question

A 3.6 µC point charge is moved within an electric field and has an electric potential energy change of 14.2 J. What is the electric potential difference before and after the charge was moved? (µC = 1.0 × 10^–6 C)

***how can I show the work for this? thank you!! :)

Answer 1

here we have to apply the subsequent formula:
\[\Large \Delta V = \frac{{\Delta U}}{q}\]
where \Delta U is the potential energy change , and \Delta V is the voltage change

Answer 2

yes:)

Answer 3

now we have to substitute the numeric values, so we get:
\[\Large \Delta V = \frac{{\Delta U}}{q} = \frac{{14.2}}{{3.6 \times {{10}^{ - 6}}}} = ...?\]

Answer 4

okay! i got 3,944,444.444! is that right?

Answer 5

no, since you have to multiply it by 10^6:

Answer 6

sorry you are right!

Answer 7

better is:
\[\Delta V = \frac{{\Delta U}}{q} = \frac{{14.2}}{{3.6 \times {{10}^{ - 6}}}} = 3.94 \times {10^6}\]

Answer 8

oops..we have to write the right unit of measure:
\[\Large \Delta V = \frac{{\Delta U}}{q} = \frac{{14.2}}{{3.6 \times {{10}^{ - 6}}}} = 3.94 \times {10^6}Volts\]

Answer 9

ohh yay!!! :D

so that is the final answer? :D

Answer 10

the question is more subtle:
if the motion of our charge is caused by the electric field itself then your answer is:
\[\Large \Delta V = - 3.94 \times {10^6}Volts\]
since positive charges are moving (under the action of an electric fields) towards point of lower electric potential

Answer 11

ah okie! so is that the final answer?

Answer 12

whereas, if the motion of our charge is due to an external force, as I suppose it is true, then our answer is:
\[\Large \Delta V = 3.94 \times {10^6}Volts\]

Answer 13

ah so that will be the final answer then? :)

Answer 14

yes! I think that your answer is the second one, namely:
\[\Delta V = 3.94 \times {10^6}Volts\]

Answer 15

yay! thank you:D

Answer 16

thank you! :)