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Tuesday, 16 August 2011

Physics GRE - #15

Two positive charges of $q$ and $2q$ coulombs are located on the x-axis at $x=0.5a$ and $1.5a$, respectively, as shown below.

There is an infinite, grounded conducting plane at $x=0$. What is the magnitude of the net force on charge q?

$\dfrac{1}{4\pi\epsilon_0}\frac{q^2}{a^2}$

$\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{2a^2}$

$\dfrac{1}{4\pi\epsilon_0}\frac{2q^2}{a^2}$

$\dfrac{1}{4\pi\epsilon_0}\frac{3q^2}{a^2}$

$\dfrac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}$

Solution :

The last choice is correct.

Recall that Coulomb's Law states the force between two point charges to be: \[F= \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\] where $q_1$ and $q_2$ are the size of the two charges, $r$ is the distance between the two charges, and all other variables are constants.

Since the conductor is grounded, image charges of $-q$ and $-2q$ are induced at $0.5a$ and $-1.5a$ respectively. These image charges pulls charge $q$ left, while the real charge $2q$ pushes charge $q$ left. Therefore we have three contributions (counted from left to right) to the force on charge $q$: \[\begin{eqnarray*}
\sum{F} & = & \frac{q}{4\pi\epsilon_0}\left(\frac{2q}{(-1.5a-0.5a)^2}+\frac{q}{(-0.5a-0.5a)^2}+\frac{2q}{(1.5a-0.5a)^2}\right) \\
& = & \frac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}.
\end{eqnarray*}\]

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What happened to the negative signs for the imaginary charges? I disagree that 7/2 is the correct factor because of this.

@7903988079171741633.0 The signs didn't come into play because we noted that the negative charges would pull 'q' left while the 2q charge would push 'q' left. Therefore, the forces add directly. The more precise answer would be \[-\frac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}\hat{x}\] where $\hat{x}$ is a unit vector in the positive $x$ direction.

I see my mistake now. However, I am still bothered why my method gives me the incorrect answer. For instance, if q is object 1 and 2q is object 2, and -q is 3, then the force on 1 by 2 (which is in the negative x direction) $\vec{F_{12}}=\frac{q}{4\pi\epsilon_0}\frac{2q}{a^2}\left(-\hat{x}\right)$ which is correct. However, when applied to object 3 $\vec{F_{13}}=\frac{q}{4\pi\epsilon_0}\frac{-q}{a^2}\left(-\hat{x}\right)=\frac{q}{4\pi\epsilon_0}\frac{q}{a^2}\hat{x}$ which isn't in the negative x direction.

@1974646916498393561.0

The version of coulomb's law you are using only gives the magnitude correctly. The vector form of coulomb's law is: \[F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}_{21}.\] This law gives the vector of the force on 1 by 2. Applying this law to object 1 and 3 gives: \[\hat{r}_{31}=+\hat{x}\implies F=\frac{1}{4\pi\epsilon_0}\frac{q\cdot(-q)}{(0.5a-(-0.5a)^2}\hat{r}_{31}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{a^2}\hat{x}\] as we wanted.

Generally, we use the non-vector form of coulomb's law first, and then add the directions based on how we think the forces should behave. It's not as proper, but its certainly quicker.

@Paul Liu Thanks Paul. I feel if I was referring to any dedicated E&M book I would of seen that vector formula. Exactly like your solution, the introductory physics textbooks I refer to avoid vectors and calculate magnitudes then indicate direction afterwards. I unfortunately prefer the formal method which I forgot how to do.

## 5 comments:

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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

The signs didn't come into play because we noted that the negative charges would pull 'q' left while the 2q charge would push 'q' left. Therefore, the forces add directly. The more precise answer would be \[-\frac{1}{4\pi\epsilon_0}\frac{7q^2}{2a^2}\hat{x}\] where $\hat{x}$ is a unit vector in the positive $x$ direction.

which is correct. However, when applied to object 3

$\vec{F_{13}}=\frac{q}{4\pi\epsilon_0}\frac{-q}{a^2}\left(-\hat{x}\right)=\frac{q}{4\pi\epsilon_0}\frac{q}{a^2}\hat{x}$ which isn't in the negative x direction.

The version of coulomb's law you are using only gives the magnitude correctly. The vector form of coulomb's law is: \[F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}_{21}.\] This law gives the vector of the force on 1 by 2. Applying this law to object 1 and 3 gives: \[\hat{r}_{31}=+\hat{x}\implies F=\frac{1}{4\pi\epsilon_0}\frac{q\cdot(-q)}{(0.5a-(-0.5a)^2}\hat{r}_{31}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{a^2}\hat{x}\] as we wanted.

Generally, we use the non-vector form of coulomb's law first, and then add the directions based on how we think the forces should behave. It's not as proper, but its certainly quicker.

Thanks Paul. I feel if I was referring to any dedicated E&M book I would of seen that vector formula. Exactly like your solution, the introductory physics textbooks I refer to avoid vectors and calculate magnitudes then indicate direction afterwards. I unfortunately prefer the formal method which I forgot how to do.

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