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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}.


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