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Monday, 22 August 2011

Physics GRE - #21

The electric potential at point P (for the ring above), which is located on the axis of symmetry a distance $x$ from the centre of the ring, is given by:

  • $\dfrac{Q}{4\pi\epsilon_0x}$
  • $\dfrac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}$
  • $\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)}$
  • $\dfrac{Qx}{4\pi\epsilon_0(R^2+x^2)^{3/2}}$
  • $\dfrac{QR}{4\pi\epsilon_0(R^2+x^2)}$

Solution :

Recall that \[V=\int{\vec{E}\cdot d\vec{l}}=\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}.\]
Since we are essentially summing up the charge over the entire ring (i.e. we are integrating around the ring with respect to $q$), we know that $int{dq}=Q$, where $Q$ is the charge of the entire ring. Therefore, \[\int{\frac{dq}{4\pi\epsilon_0|\vec{r}|}}=\frac{Q}{4\pi\epsilon_0|\vec{r}|}=\frac{Q}{4\pi\epsilon_0\sqrt{R^2+x^2}}\] where $|\vec{r}|=\sqrt{R^2+x^2}$ as indicated in the diagram below (except they used $z$ for $x$).


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