A parallel-plate capacitor with plates of area 590 cm2 and is connected across the terminals of...

A parallel-plate capacitor with plates of area 590 cm2 and is connected across the terminals of a battery. After some time has passed, the capacitor is disconnected from the battery. When the plates are then moved 0.48 cm farther apart, the charge on each plate remains constant but the potential difference between the plates increases by 100 V.

(a) What is the magnitude of the charge on each plate?

(b) Do you expect the energy stored in the capacitor to increase, decrease, or remain constant as the plates are moved this way? Explain your answer.
(c) Support your answer to Part (b) by determining the change in stored energy in the capacitor due to the movement of the plates.

Expert Solution

Let the initia; seperation between the plate is then capacitance will be

Initial voltage will be

Potential after the increment in separation

$V_f=V_i+100=\frac{Q(d+0.48\times10^{-2})}{\epsilon_0 A}\\ \Rightarrow \frac{Qd}{\epsilon_0 A}+100=\frac{Q(d+0.48\times10^{-2})}{\epsilon_0 A}\Rightarrow 100=\frac{Q\times0.48\times10^{-2}}{\epsilon_0 A}\\ \Rightarrow Q=\frac{100\times\epsilon_0 A}{0.48\times10^{-2}}=\frac{100\times8.85\times10^{-12}\times 590\times10^{-4}}{0.48\times10^{-2}}=10.878$ nC$$

(b) Energy stored in the capacitor will increase as by separating the plate, external work is needed to be done on the system (Capacitor) and that work will be stored in the form of electrostatic energy.

(c) Initial, final and change in energy in the capacitor will be

$U_i=\frac{1}{2}QV_i\\ U_f=\frac{1}{2}Q\times(V_i+100)\\ \Delta U=U_f-U_i=\frac{1}{2}Q\times(V_i+100)-\frac{1}{2}QV_i=\frac{1}{2}Q\times100=50Q\\ \Delta U=5.439\times10^{-7}$ J$$

genius_generous answered 1 year ago

A parallel-plate capacitor having plates 6.00cmapart is connected across the terminals of a 12.0Vbattery. A. Being...

A parallel-plate capacitor having plates 6.00cmapart is connected across the terminals of a 12.0Vbattery. A. Being as quantitative as you can, describe the location of the equipotential surface that is at a potential of 6.00V relative to the potential of the negative plate. Avoid the edges of the plates. d= B. Do the same for the equipotential surface that is at 2.00V relative to the negative plate. d= C. What is the potential gradient between the plates? E=

The plates of an air-filled parallel-plate capacitor with a plate area of 15.0 cm2 and a...

The plates of an air-filled parallel-plate capacitor with a plate area of 15.0 cm2 and a separation of 8.95 mm are charged to a 170-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with κ = 6.5 is inserted between the plates of the capacitor. (a) What is the charge on the capacitor before and after the dielectric is inserted? Qi = C Qf = C (b) What is the capacitance of the capacitor after...

A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 2.00 mm with...

A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 2.00 mm with air between them. The capacitor stores a charge of 500 pC.(a) What is the potential difference across the plates of the capacitor?1 V(b) What is the magnitude of the uniform electric field in the region between the plates?

The parallel-plates in a capacitor, with a plate area 1.2×10−3 cm2 and an air-filled separation of...

The parallel-plates in a capacitor, with a plate area 1.2×10−3 cm2 and an air-filled separation of 10 mm, are charged by a 12 V battery. They are then disconnected from the battery and pushed together (without discharge) to a separation of 3.5 mm. (a) Find the initial potential difference between the plates and the initial stored energy. (6 points) (b) Find the final potential difference between the plates and the final stored energy. (6 points) (c) How much work is...

A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and a...

A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and a separation of 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64E-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

A parallel plate air capacitor with no dielectric between the plates is connected to a constant...

A parallel plate air capacitor with no dielectric between the plates is connected to a constant voltage source. How would the capacitance and the charge change if a dielectric of dielectric constant K=2 is inserted between the plates. C0 and Q0 are the capacitance and charge of the capacitor before the introduction of the dielectric.

A parallel-plate capacitor with circular plates and a capacitance of 10.3 μF is connected to a...

A parallel-plate capacitor with circular plates and a capacitance of 10.3 μF is connected to a battery which provides a voltage of 11.2 V . What is the charge on each plate? How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was doubled without changing their...

An empty parallel plate capacitor is connected between the terminals of a 10.7-V battery and charged...

An empty parallel plate capacitor is connected between the terminals of a 10.7-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d =...

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 6.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . a) Find the energy U1 of the dielectric-filled capacitor. Express your answer numerically in joules. b) The dielectric plate is now slowly pulled out of the capacitor, which remains connected...

A dielectric-filled parallel-plate capacitor has plate area A = 25.0 cm2 , plate separation d =...

A dielectric-filled parallel-plate capacitor has plate area A = 25.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 2.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . PART A: U1 was found: U1 = 3.84×10−10 J PART B: The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the...

Question