7.5 Equipotential surfaces and conductors  (Page 3/9)

Calculating equipotential lines

You have seen the equipotential lines of a point charge in [link] . How do we calculate them? For example, if we have a $\text{+}10\text{-nC}$ charge at the origin, what are the equipotential surfaces at which the potential is (a) 100 V, (b) 50 V, (c) 20 V, and (d) 10 V?

Strategy

Set the equation for the potential of a point charge equal to a constant and solve for the remaining variable(s). Then calculate values as needed.

Solution

In $V=k\frac{q}{r}$ , let V be a constant. The only remaining variable is r ; hence, $r=k\frac{q}{V}=\text{constant}$ . Thus, the equipotential surfaces are spheres about the origin. Their locations are:

1. $r=k\frac{q}{V}=\left(8.99\times {10}^9{\text{Nm}}^2{\text{/C}}^2\right)\frac{\left(10\times {10}^{\mathrm{-9}}\text{C}\right)}{100\text{V}}=0.90\text{m}$ ;
2. $r=k\frac{q}{V}=\left(8.99\times {10}^9{\text{Nm}}^2{\text{/C}}^2\right)\frac{\left(10\times {10}^{\mathrm{-9}}\text{C}\right)}{50\text{V}}=1.8\text{m}$ ;
3. $r=k\frac{q}{V}=\left(8.99\times {10}^9{\text{Nm}}^2{\text{/C}}^2\right)\frac{\left(10\times {10}^{\mathrm{-9}}\text{C}\right)}{20\text{V}}=4.5\text{m}$ ;
4. $r=k\frac{q}{V}=\left(8.99\times {10}^9{\text{Nm}}^2{\text{/C}}^2\right)\frac{\left(10\times {10}^{\mathrm{-9}}\text{C}\right)}{10\text{V}}=9.0\text{m}$ .

Significance

This means that equipotential surfaces around a point charge are spheres of constant radius, as shown earlier, with well-defined locations.

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Potential difference between oppositely charged parallel plates

Two large conducting plates carry equal and opposite charges, with a surface charge density $\sigma$ of magnitude $6.81\times {10}^{\mathrm{-7}}{\text{C/m}}^2,$ as shown in [link] . The separation between the plates is $l=6.50\text{mm}$ . (a) What is the electric field between the plates? (b) What is the potential difference between the plates? (c) What is the distance between equipotential planes which differ by 100 V?

Strategy

(a) Since the plates are described as “large” and the distance between them is not, we will approximate each of them as an infinite plane, and apply the result from Gauss’s law in the previous chapter.

(b) Use $\text{Δ}{V}_{AB}=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ .

(c) Since the electric field is constant, find the ratio of 100 V to the total potential difference; then calculate this fraction of the distance.

Solution

1. The electric field is directed from the positive to the negative plate as shown in the figure, and its magnitude is given by
   
   $E=\frac{\sigma }{{\epsilon }_0}=\frac{6.81\times {10}^{\mathrm{-7}}{\text{C/m}}^2}{8.85\times {10}^{\mathrm{-12}}{\text{C}}^2\text{/}\text{N}·{\text{m}}^2}=7.69\times {10}^4\text{V/m}\text{.}$
2. To find the potential difference $\text{Δ}V$ between the plates, we use a path from the negative to the positive plate that is directed against the field. The displacement vector $d\overrightarrow{\text{l}}$ and the electric field $\overrightarrow{\text{E}}$ are antiparallel so $\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}=\text{−}Edl.$ The potential difference between the positive plate and the negative plate is then
   
   $\text{Δ}V=\text{−}{\displaystyle \int E·dl=E{\displaystyle \int dl=El}}=(7.69\times {10}^4\text{V/m})(6.50\times {10}^{\mathrm{-3}}\text{m})=500\text{V}\text{.}$
3. The total potential difference is 500 V, so 1/5 of the distance between the plates will be the distance between 100-V potential differences. The distance between the plates is 6.5 mm, so there will be 1.3 mm between 100-V potential differences.

Significance

You have now seen a numerical calculation of the locations of equipotentials between two charged parallel plates.

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