Question

3. A 5.0 µC point charge is moved within an electric field and has an electric potential energy change of 10.0 J. What is the electric potential difference before and after the charge was moved?

Answer 1

here we have to apply the subsequent formula:
\[\Large \Delta U = q\Delta V\]

Answer 2

where:
\[\Delta U\] is the change in energy
and:
\[\Delta V\]
is the requested potential difference

Answer 3

q is the cahrge, of course

Answer 4

charge*

Answer 5

so, we have:
\[\Large \Delta V = \frac{{\Delta U}}{q} = ...Volts\]

Answer 6

Okay, so all of the things needed in the equation are in the question?

So, q = 5?
ΔU = 10?
Am I doing that right?

Answer 7

q= 5*10^(-6) Coulombs

Answer 8

Oh, okay. So then ΔV = 10? Or..

Answer 9

\[\Large \Delta V = \frac{{\Delta U}}{q} = \frac{{10}}{{5 \times {{10}^{ - 6}}}} = \frac{{10}}{5} \times {10^6} = ...Volts\]

Answer 10

@jaysabelle

Answer 11

So it would be \[2x10^{6}\] = volts?

Answer 12

@Michele_Laino

Answer 13

that's right!

Answer 14

your answer is correct!

Answer 15

Really? Thank you!