i need help in solving this problem, please help.
Four identical charges (+3.0 ìC each) are brought from infinity and fixed to a straight line. Each charge is 0.47 m from the next. Determine the electric potential energy of this group.
4 answers
i used this equation but i didn't get the correct answer.
What does your " ì " symobol mean in front of the C? Let Q be the charge of each.
Imagine bringing the charges together one at a time. The work required to bring the second charge within a distance a of the first is kQ/a. The third charge in the line requires kQ/a + kQ/2a. Add to that the work required for the fourth charge and that will be the potential energy of the configuration.
In a previous post (now removed) I said to reverse the sign of the work done. That was incorrect.
Imagine bringing the charges together one at a time. The work required to bring the second charge within a distance a of the first is kQ/a. The third charge in the line requires kQ/a + kQ/2a. Add to that the work required for the fourth charge and that will be the potential energy of the configuration.
In a previous post (now removed) I said to reverse the sign of the work done. That was incorrect.
that symbol means micro, i copied and pasted it off an internet post.
I also forgot to square Q in my previous answer. I forgot that electric potential has to be multiplied by Q to get energy.
The answer I get now
P.E. = (4 1/3) k Q^2/a
Which computes to
k*(4 1/3)*(3*10^-6 C)^2/(0.47 m)
Using k = 8.99*10^9 N/(m^2C^2, that is
0.746 J
The answer I get now
P.E. = (4 1/3) k Q^2/a
Which computes to
k*(4 1/3)*(3*10^-6 C)^2/(0.47 m)
Using k = 8.99*10^9 N/(m^2C^2, that is
0.746 J
1 answer