The electric field between the plates of an isolated air-spaced parallel- plate capacitor is E. What is the field between the plates after immersing the capacitor in a liquid of relative permitivity 10?
Correct answer: D. E/10
- A. E√10
- B. E/√10
- C. 10E
- D. E/10
Explanation
The relationship between the electric field in the absence of the dielectric (E0) and the electric field with the dielectric (E) is given by:E = E0 / εrwhere εr is the relative permittivity of the medium.In this case, the relative permittivity of the liquid is given as 10. Therefore, the fiel used between the plates after immersing the capacitor in the liquid is:E' = E / εr = E / 10So, the correct answer is option 4E / 10SO, ELECTRIC FIELD WILL DECREASE BY A FACTOR 10.
Last updated
About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
A 100µF capacitor with a 12V source in series having frequency 50Hz will offer a capacitive reactance of about______________?
A 18.0 V battery is connected to a capacitor, resulting in 27.0 µC of charge stored on the capacitor. How much energy is stored in the capacitor?
A 2uF capacitor is connected to a 10 V dc supply, if 'Q' represents the charge on each plate and 'E' the energy stored on each plate, which of the following represents the values of Q and E?
A battery is permanently connected to a parallel plate capacitor and the energy stored is x joules. When one plate is moved so that the separation of the plate is doubled, the energy now stored in the joule is:
A capacitor of capacitance 30µF is charged by a constant current of 10mA. If initially, the capacitor was uncharged what is the time taken for the potential difference across the capacitor to reach 300V ?