Chapter 17: Q52P (page 473)
Question: (I) 650 V is applied to a 2800-pF capacitor. How much energy is stored?
Short Answer
The stored energy in the capacitor is \(5.9 \times {10^{ - 4}}{\rm{J}}\).
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 17: Q52P (page 473)
Question: (I) 650 V is applied to a 2800-pF capacitor. How much energy is stored?
The stored energy in the capacitor is \(5.9 \times {10^{ - 4}}{\rm{J}}\).
All the tools & learning materials you need for study success - in one app.
Get started for free
Question: (I) A cardiac defibrillator is used to shock a heart that is beating erratically. A capacitor in this device is charged to 5.0 kV and stores 1200 J of energy. What is its capacitance?
Question: (I) What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 2.8 mm of mica?
(II) (a) What is the electric potential \({\bf{2}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{m}}\) away from a proton (charge +e)? (b) What is the electric potential energy of a system that consists of two protons \({\bf{2}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{m}}\) apart—as might occur inside a typical nucleus?
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?
When the proton and electron in MisConceptual Question 6 strike the opposite plate, which one has more kinetic energy?
(a) The proton.
(b) The electron.
(c) Both acquire the same kinetic energy.
(d) Neither—there is no change in kinetic energy.
(e) They both acquire the same kinetic energy but with opposite signs.
What do you think about this solution?
We value your feedback to improve our textbook solutions.