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- Thread starter lorx99
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TSny

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Shouldn'tUf = k(7*(-5) + 7(-4) + (-5)*(-4))/0.1 = -4.3*10^-4

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gneill

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TSny

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Note that in the choices of answers, the symbol* k* represents Coulomb's constant, not* kilo*.

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I aciddently left out that i multiplied by 10^-6 for the product of q's.

But the answer is right.

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gneill

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I get a different result on the order of a few Joules. Maybe check your arithmetic?I aciddently left out that i multiplied by 10^-6 for the product of q's.

But the answer is right.

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Thanks, i entered the E-6 wrong! answer is -150e-12I get a different result on the order of a few Joules. Maybe check your arithmetic?

- #8

gneill

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##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

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Hi, gneill. Apparently they don't want you to substitue a value for ##k##. Thus,##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)##

##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##.

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gneill

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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.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##.

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TSny

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Yes, it threw me off at first. In the problem statement, it says, "answer in terms of k = 1/(4πεHmm. 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.

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