A parallel plate capacitor is charged by a battery of potential V and then disconnected. Now the distance between the plates of capacitor is doubled then the electric field between the plates of capacitor will beChoose answer: Doubled Halved Quadrupled Remain same
Question
A parallel plate capacitor is charged by a battery of potential V and then disconnected. Now the distance between the plates of capacitor is doubled then the electric field between the plates of capacitor will beChoose answer: Doubled Halved Quadrupled Remain same
Solution
The electric field (E) between the plates of a capacitor is given by the equation E = V/d, where V is the voltage across the capacitor and d is the distance between the plates.
When the distance between the plates is doubled, the new electric field E' is given by E' = V/(2d).
Therefore, the new electric field is half the original electric field. So, the electric field between the plates of the capacitor will be halved.
Similar Questions
A parallel plate capacitor is connected to a battery of potential V. Now if the distance between the plates of capacitor is doubled (keeping battery connected) then the energy stored in the capacitor will beChoose answer: Doubled Halved Quadrupled Remain same
A parallel plate capacitor is charged so that a total charge of +Q ends up on the positive plate, and -Q on the negative plate.The capacitor is then disconnected from rest of the circuit. Next, I push the two plates slightly closer together. Which statement is true about the electric field and the voltage difference between the plates?
A parallel plate capacitor initially has a voltage of 400V, and stays connected to a battery. If the plate spacing is doubled, what happens? 1. the voltage decreases 2. the voltage increases 3. the charge decreases 4. the charge increases 5. both voltage and charge change
If I double the distance between the two plates of a parallel plate capacitor while holding the voltage constant, what happens to the strength of the electric field? Question 3Select one:a.It is reduced to half it's original valueb.It doublesc.It is reduced but not by one half.d.There is not enough information to say.
A parallel plate capacitor of capacitance C with plates at separation d, has charge Q. The electric field at a distance r from positive plate between the plates isChoose answer:
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.