Question

Assertion : A parallel plate capacitor is first charged and then distance between the plates is increased. In this process, electric field between the plates remains the same, while potential difference gets decreased. Reason: E=q/(Aepsilon_0) and V=q/(Adepsilon_0) Since q, remain same, E will remain same while V will decrease.

Answer 1

Answer:

367rykvftih

Explanation:

frieuf5ui

645te5

446yfrfgh

rdbb7d5afum

5dvubino7v6c

a

Answer 2

Both Assertion and Reason are false.

• Electric field will remain same while the voltage will increase.

• The reason for Electric field to remain same is right.
• But as E=V/d, so if E remains the same and d increases so V has to increase in order to maintain the equality.
• It can also be stated as V can be written as :

                   V=Qd/εA      (as, Q=CV and C=εA/d)

As d increases so V will also increases.