Answered: A parallel plate capacitor is charged… | bartleby

Related questions

Question

A parallel plate capacitor is charged by a battery.
After charging the capacitor, battery is disconnected
and distance between the plates is decreased then
which of the following statement is correct ?
(1) electric field does not remain constant
(2) potential difference is increased
(3) the capacitance decreases
(4) the stored energy decreases

Expert Solution

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A parallel-plate capacitor is fully charged by a battery. While still connected to the battery, the distance between the plates increases. What happens to the charge on the capacitor? Answers: decreases increases stays the same
A parallel-plate capacitor is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. Which of the following is true? Select one. a. D increases, E increases b. D remains the same, E increases c. D decreases, E increases d. D increases, E remains the same e. D remains the same, E remains the same f. D decreases, E remains the same g. D increases, E decreases h. D remains the same, E decreasesi. D decreases, E decreases
Chapter 26 A parallel-plate capacitor has plates of area 0.2 m? and a separation of 0.9 cm. A battery charges the plates to a potential difference of 100 V and is then disconnected. A dielectric slab of thickness 9 mm and dielectric constant 2 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? (c) What is the free charge q before the slab is inserted? and (d) What is the free charge q after the slab is inserted? (e) What is the magnitude of the electric field in the space between the plates and dielectric before inserting the dielectric? (f) What is the magnitude of the electric field in the space between the plates and dielectric after inserting the dielectric? (g) With the slab in place, what is the potential difference across the plates?
A parallel-plate capacitor is fully charged by a battery. Now the battery is disconnected, and the distance between the plates increases. What happens to the electric field in the capacitor? Answers: stays the same increases decreases
The volume enclosed by the plates of a parallel plate capacitor is given by the equation Volume = Ad. If one doubles the area, A, of the plates and reduces the plate separation, d, by a factor of two so that the volume remains the same, the capacitance will do which of the following? decrease by a factor of 4 O increase by a factor of 4 increase by a factor of 2 decrease by a factor of 2 O remain constant
An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is AV and the magnitude of the electric field between the plates is E. If the battery is disconnected, and then the distance between the plates is increased, O AV increases, E remains the same O AV remains the same, E remains the same O AV increases, E increases O AV decreases, E decreases O AV decreases, E remains the same O AV remains the same, E increases O AV decreases, E increases O AV increases, E decreases O AV remains the same, E decreases
An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is AV and the magnitude of the electric field between the plates is E. If the battery is disconnected, and then a dielectric is inserted between the plates O AV remains the same, E increases O AV increases, E increases O AV remains the same, E remains the same O AV decreases, E increases O AV decreases, E decreases O AV remains the same, E decreases O AV decreases, E remains the same O AV increases, E decreases O AV increases, E remains the same