• Support PF! Buy your school textbooks, materials and every day products Here!

Are the answer choices wrong? (electric potential energy)

  • Thread starter lorx99
  • Start date
  • #1
21
0

Homework Statement


upload_2018-10-7_17-16-10.png


Homework Equations


U=kq1q2/r

The Attempt at a Solution


W = changeU = Uf-Uo

Uf = k(7*(-5) + 7(-4) + (-5)*(-4))/0.1 = -4.3*10^-4
Ui= k((7*(-4))/0.1= -2.8*10^-4

Uf-Ui = -1.5*10^-4k J
 

Attachments

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,411
2,846
Uf = k(7*(-5) + 7(-4) + (-5)*(-4))/0.1 = -4.3*10^-4
Shouldn't k appear in the expression on the right side? How did you get the power of -4? Did you take into account that the charges are in micro Coulombs? Otherwise, your approach looks right.
 
  • #3
gneill
Mentor
20,792
2,770
The provided answer choices seem out of line for the given problem statement, but your calculated answer is also rather suspect. How did you determine the order of magnitude of the results? What value did you use for ##k##?
 
  • #4
TSny
Homework Helper
Gold Member
12,411
2,846
Note that in the choices of answers, the symbol k represents Coulomb's constant, not kilo.
 
  • #5
21
0
The provided answer choices seem out of line for the given problem statement, but your calculated answer is also rather suspect. How did you determine the order of magnitude of the results? What value did you use for ##k##?
I aciddently left out that i multiplied by 10^-6 for the product of q's.

But the answer is right.
 
  • #6
gneill
Mentor
20,792
2,770
I aciddently left out that i multiplied by 10^-6 for the product of q's.

But the answer is right.
I get a different result on the order of a few Joules. Maybe check your arithmetic?
 
  • #7
21
0
I get a different result on the order of a few Joules. Maybe check your arithmetic?
Thanks, i entered the E-6 wrong! answer is -150e-12
 
  • #8
gneill
Mentor
20,792
2,770
Okay, let's take a look at the initial electric potential energy of the original configuration comprised of the two first charges:

##q_1 = 7~μC##
##q_2 = -4 μC##
##D = 0.1~m##

##U_o = k\frac{q_1 q_2}{D}##
##U_o = 8.988 \times 10^9~\frac{V~m}{C}\left( \frac{7\times10^-6~C \cdot (-4\times 10^-6~C)}{0.1~m} \right)##
I find that:

##U_o = -2.52~J## or, ##U_o = -2.52~\times 10^{-3}~kJ##

So we can expect answers to be on the order of ##10^1## Joules
 
  • #9
TSny
Homework Helper
Gold Member
12,411
2,846
##U_o = 8.988 \times 10^9~\frac{V~m}{C}\left( \frac{7\times10^-6~C \cdot (-4\times 10^-6~C)}{0.1~m} \right)##
Hi, gneill. Apparently they don't want you to substitue a value for ##k##. Thus,

##U_o = k \left( \frac{7\times10^-6~C \cdot (-4\times 10^-6~C)}{0.1~m} \right) = k~ \left(-280 \times10^{-12} \, C^2/m \right) = -280 \times10^{-12}~ k~ J##.

Here, the ##k## is Coulomb's constant (even in the final expression). The units for ##k## have been absorbed into ##J## in the last step. This is an awkward way to express the answer, but I guess they didn't want the student to bother with looking up the value of ##k##.
 
  • #10
gneill
Mentor
20,792
2,770
Hi, gneill. Apparently they don't want you to substitue a value for ##k##. Thus,

##U_o = k \left( \frac{7\times10^-6~C \cdot (-4\times 10^-6~C)}{0.1~m} \right) = k~ \left(-280 \times10^{-12} \, C^2/m \right) = -280 \times10^{-12}~ k~ J##.

Here, the ##k## is Coulomb's constant (even in the final expression). The units for ##k## have been absorbed into ##J## in the last step. This is an awkward way to express the answer, but I guess they didn't want the student to bother with looking up the value of ##k##.
Hmm. Okay, I wasn't expecting that. When I see kJ I immediately think kilo-Joules. It seems to me a bit odd to expect students to know that they need not invoke the relevant constant values.
 
  • #11
TSny
Homework Helper
Gold Member
12,411
2,846
Hmm. Okay, I wasn't expecting that. When I see kJ I immediately think kilo-Joules. It seems to me a bit odd to expect students to know that they need not invoke the relevant constant values.
Yes, it threw me off at first. In the problem statement, it says, "answer in terms of k = 1/(4πε0)." It could have been clearer as what was meant here.
 

Related Threads on Are the answer choices wrong? (electric potential energy)

Replies
15
Views
3K
  • Last Post
Replies
1
Views
866
Replies
0
Views
2K
Replies
2
Views
11K
Replies
5
Views
5K
Replies
2
Views
2K
  • Last Post
Replies
11
Views
5K
Replies
11
Views
1K
Replies
8
Views
262
Top