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Which of the following do not affect capacitance?

(a) Area of the plates.

(b) Separation of the plates.

(c) Material between the plates.

(d) Charge on the plates.

(e) Energy stored in the capacitor.

Short Answer

Expert verified

(d) Charge on the plates.

(e) Energy stored in the capacitor.

Step by step solution

01

Understanding the capacitance of the capacitor

Capacitance establishes an inverse relation with the potential difference and a direct relation with the charge of the plates. In the parallel plates, the capacitance shows an inverse relation with the separation distance of the plates.

02

Determination of the factors that do not affect the capacitance

The capacitance can be expressed as:

\(C = \frac{Q}{V}\)

Here, Qis the charge andVis the potential difference. The above equation of capacitance gives the constant ratio value.

The variation in the charge will cause the potential difference to vary. There is no change in the capacitance.

For the parallel plate’s capacitor, thecapacitance can be expressed as:

\(C = { \in _0}\frac{A}{d}\)

Here, A is the area of the plates, d is the separation distance, and\({ \in _0}\)is the permittivity of free space.

The potential energy stored by the capacitor can be expressed as:

\(\begin{aligned}E &= \frac{1}{2}C{V^2}\\C &= \frac{{2E}}{{{V^2}}}\end{aligned}\)

The variation in the potential energy will cause the charge of the parallel plates and the difference in the potential between the plates to vary. This will not change the capacitance of the plates.

Thus, the capacitance is affected by the plates' area, the separation between the plates, and the dielectric material between the two plates.

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Most popular questions from this chapter

A parallel-plate capacitor with plate area\({\bf{A = 2}}{\bf{.0}}\;{{\bf{m}}{\bf{2}}}\)and plate separation\({\bf{d = 3}}{\bf{.0}}\;{\bf{mm}}\)is connected to a 35-V battery (Fig. 17–51a).

(a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor.

(b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength K =3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17–51b). What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor?\(\left( {{\bf{1}}\;{\bf{byte = 8}}\;{\bf{bits}}{\bf{.}}} \right)\)

FIGURE 17-51 Problem 96

Two identical positive charges are placed near each other. At the point halfway between the two charges,

(a) the electric field is zero and the potential is positive.

(b) the electric field is zero and the potential is zero.

(c) the electric field is not zero and the potential is positive.

(d) the electric field is not zero and the potential is zero.

(e) None of these statements is true.

A dielectric is pulled out from between the plates of a capacitor which remains connected to a battery. What changes occur to (a) the capacitance, (b) the charge on the plates, (c) the potential difference, (d) the energy stored in the capacitor, and (e) the electric field? Explain your answers.

A lightning flash transfers 4.0 C of charge and 5.2 MJ of energy to the Earth.

(a) Across what potential difference did it travel?

(b) How much water could this boil and vaporize, starting from room temperature? (See also Chapter 14.)

If two points are at the same potential, does this mean that no net work is done in moving a test charge from one point to the other? Does this imply that no force must be exerted? Explain.

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