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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

Expert verified

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

Step by step solution

01

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, \(s\)
02

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.

03

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 \(q = Q\left( {\frac{s}{{2R - s - 2t}}} \right)\)

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Most popular questions from this chapter

A capacitor consists of two large metal disks of radius 1.1 m placed parallel to each other, a distance of 1.2 mm apart. The capacitor is charged up to have an increasing amount of charge +Q on one disk and −Q on the other. At about what value of Q does a spark appear between the disks?

If the magnitude of the electric field in air exceeds roughly 3 × 106 N/C, the air brake down and a spark form. For a two-disk capacitor of radius 47 cm with a gap of 1 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)?

A plastic rod 1.7mlong is rubbed all over with wool, and acquires a charge of-2×10-8C(Figure 15.52). We choose the center of the rod to be the origin of our coordinate system, with the x axis extending to the right, the y axis extending up, and the z axis out of the page. In order to calculate the electric field at locationA=<07,0,0>, we divide the rod into eight pieces, and approximate each piece as a point charge located at the center of the piece.

(a) What is the length of one of these pieces? (b) What is the location of the center of piece number 3? (c) How much charge is on piece number? (Remember that the charge is negative.) (d) Approximating piece 3as a point charge, what is the electric field at location A due only to piece 3? (e) To get the net electric field at location A, we would need to calculatedue to each of the eight pieces, and add up these contributions. If we did that, which arrow (a–h) would best represent the direction of the net electric field at location A?

For a disk of radius R=20cm and Q=6×10-6C, calculate the electric field 2 mm from the center of the disk using all three equations:

role="math" localid="1656928965291" E=(Q/A)2ε0[1-z(R2+z)1/2]

E≈Q/A2e0[1-zR],andE≈Q/A2e0

How good are the approximate equations at this distance? For the same disk, calculate E at a distance of 5 cm (50 mm) using all three equations. How good are the approximate equations at this distance?

A thin rod lies on the x axis with one end atand the other end at-A, as shown in Figure 15.51. A charge of-Q
is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <0,Y,0>due to the rod. Following the procedure discussed in this chapter, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram

Answer using the variables x,y,dx,A,Qas appropriate. Remember that the rod has charge-Q. (a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod? (b) What is the amount of chargedQon the small piece of lengthdx? (c) What is the vector from this source to the observation location? (d) What is the distance from this source to the observation location? (e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

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