Question

A capacitor made of two parallel uniformly charged circular metal disks carries a charge of +Q and −Q on the inner surfaces of the plates and very small amounts of charge +q and −q on the outer surfaces of the plates. Each plate has a radius R and thickness t, and the gap distance between the plates is s. How much charge q is on the outside surface of the positive disk, in terms of Q?

Short answer

The charge on the outer surface of plate/disk is \(q = Q\left( {\frac{s}{{2R - s - 2t}}} \right)\)

Step-by-step solution

Identification of given data

The given data can be listed below,

• The radius of the plates is,\(R\)
• The thickness of the plates is,\(t\)
• The gap between the plates is, \(s\)

Concept/Significance of parallel plate capacitor

It's a device that allows it to raise the capacitance of a conducting plate without affecting the size of the plate.

Determination of charge q is on the outside surface of the positive disk.

The electric fields of each plate on the middle of the upper plate (point x in the figure) are

\(\begin{aligned}{E_{ + q}} &= - \frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{{\sqrt {{R^2} + {{\left( {t{\rm{/}}2} \right)}^2}} }}} \right)\\{E_{ + Q}} &= \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{{\sqrt {{R^2} + {{\left( {t{\rm{/}}2} \right)}^2}} }}} \right)\\{E_{ - Q}} &= - \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{{\sqrt {{R^2} + {{\left( {s + 3t{\rm{/}}2} \right)}^2}} }}} \right)\\{E_{ - q}} &= - \frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{{\sqrt {{R^2} + {{\left( {s + 3t{\rm{/}}2} \right)}^2}} }}} \right)\end{aligned}\)

Here, \({E_{ + q}}\) is the electric field due to \( + q\) charge, t is the thickness of each plate, s is the gap between the plates, R is the radius of each plate and A is the area of each plate.

From the conservation of energy, the total electric field is given by,

\({E_{ + q}} + {E_{ + Q}} + {E_{ - Q}} + {E_{ - q}} = 0\)

The radius of the plate is much larger than the gap and thickness so these quantities can be neglected. Substitute all the values in the above,

\(\begin{aligned}0 &= - \frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{R}} \right) + \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{R}} \right) - \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{R}} \right) - \frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{R}} \right)\\\frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{R}} \right) + \frac{{q{\rm{/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{R}} \right) &= \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{t{\rm{/}}2}}{R}} \right) - \frac{{{\rm{Q/}}A}}{{2{\varepsilon _0}}}\left( {1 - \frac{{s + t{\rm{/}}2}}{R}} \right)\\q\left( {2 - \frac{{s + 2t}}{R}} \right) = Q\left( {\frac{s}{R}} \right)\\q &= Q\left( {\frac{s}{{2R - s - 2t}}} \right)\end{aligned}\)

Thus, the charge on the outer surface of the plate/disk is \(q = Q\left( {\frac{s}{{2R - s - 2t}}} \right)\)