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Chapter 17: Q15Q (page 473)

If the voltage across a fixed capacitor is doubled, the amount of energy it stores (a) doubles; (b) is halved; (c) is quadrupled; (d) is unaffected; (e) none of these. Explain.

Short Answer

Expert verified

The correct answer is option (c) is quadrupled

Step by step solution

01

Understanding the capacitance of a capacitor

The capacitance of a capacitor does get affected by the change in the voltage across the capacitor and energy stored in it. It depends on the separation between the plates and area of each plate.

02

Evaluation of the change in amount of energy

The relation of energy stored in a capacitor is given by,

\(E = \frac{1}{2}C{V^2}\)

Here, Eis the energy stored, Cis the capacitance and Vis the voltage.

In the above relation, the energy stored in the capacitor relies on the voltage supplied, which means that it increases with a rise in voltage. Here, if the voltage is doubled, the energy will become quadruple because of the square of the voltage.

Thus, option (c) is the correct answer.

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Most popular questions from this chapter

Draw in a few equipotential lines in Fig. 16–32b and c.

FIGURE 16-32

(III) In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus) in a circular orbit of radius \({\bf{0}}{\bf{.53 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\). (a) What is the electric potential at the electron’s orbit due to the proton? (b) What is the kinetic energy of the electron? (c) What is the total energy of the electron in its orbit? (d) What is the ionization energy— that is, the energy required to remove the electron from the atom and take it to \({\bf{r = }}\infty \), at rest? Express the results of parts (b), (c), and (d) in joules and eV.

Question: A \({\bf{3}}{\bf{.4}}\;{\bf{\mu C}}\) and a \({\bf{ - 2}}{\bf{.6}}\;{\bf{\mu C}}\) charge are placed 2.5 cm apart. At what points along the line joining them is (a) the electric field zero, and (b) the electric potential zero?

(III) How much voltage must be used to accelerate a proton (radius \({\bf{1}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{m}}\)) so that it has sufficient energy to just “touch” a silicon nucleus? A silicon nucleus has a charge of \( + 14e\), and its radius is about \({\bf{3}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{m}}\). Assume the potential is that for point charges.

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.
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