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.
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.
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
To determine the electric potential energy of this group of charges, you need to follow these steps:
Step 1: Determine the electric potential energy between two charges
The electric potential energy between two point charges can be calculated using the formula:
PE = k * (q1 * q2) / r
Where:
- PE is the electric potential energy
- k is the Coulomb's constant, approximately 8.99 × 10^9 N m^2 / C^2
- q1 and q2 are the magnitudes of the charges
- r is the distance between the charges
Step 2: Calculate the electric potential energy for each pair of charges
Since you have four identical charges, you will have three pairs of charges. Calculate the electric potential energy for each pair of charges separately.
Step 3: Calculate the total electric potential energy
To find the total electric potential energy of the group, add together the electric potential energy of each pair of charges.
Let's go through the calculations:
Step 1: Determine the electric potential energy between two charges (q1, q2 = 3.0 µC each and r = 0.47 m)
PE1 = (8.99 × 10^9 N m^2 / C^2) * [(3.0 × 10^-6 C) * (3.0 × 10^-6 C)] / 0.47 m
= 162.42 J
Step 2: Calculate the electric potential energy for each pair of charges
For the first pair of charges, the electric potential energy is 162.42 J.
For the second pair of charges, which has the same distance, the electric potential energy will also be 162.42 J.
For the third pair of charges, since the distance is the same, the electric potential energy will still be 162.42 J.
Step 3: Calculate the total electric potential energy
Total Electric Potential Energy = PE1 + PE2 + PE3
= 162.42 J + 162.42 J + 162.42 J
= 487.26 J
Therefore, the electric potential energy of this group of charges is 487.26 J.