Question

A parallel plate capacitor is connected to a battery for charging. after some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled.
which of the following statements are true ?

a) the charge on the plates does not change.

b) the capacitance doubles.

c) the potential difference of the battery is halved.

d) the electric field between the plates is halved.

e) the potential difference across the plates does not change.

f) none of the above.

Answer 1

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

Answer 2

what does that mean !

Answer 3

when d increase, capacitance decrease ...

Answer 4

so !! which answer is the right one !

Answer 5

you guess!!

Answer 6

f !!

Answer 7

or D ! right !

Answer 8

i think it's d !

Answer 9

it don't think D

Answer 10

what !

Answer 11

so which answer !

Answer 12

guess

Answer 13

so it's f !

Answer 14

there might be some correct.

Answer 15

its D.

Answer 16

might be ... but i think e might be correct too.

Answer 17

hey ! i want one answer for god sake :D u both are driving me crazy !

Answer 18

i haven't checked the possibility of d)


Oo.. V = Ed
=> E = V/d
=> E = V/2d
guess this is right too.

Answer 19

V is directly proportional to Q and the increase in d will cause the P.D to decrease as
v=kq/r

so D is the correct one

Answer 20

still, it seems the potential across the place is same.

Answer 21

LOL ... i got confused, you can do with D

Answer 22

:D ! finally :D

Answer 23

well sorry for the mess.

Answer 24

no it's ok :D ! ThnQQ !

Answer 25

@experimentX , the electric field is sigma/ (epsilon nought) between 2 plates of capacitor. How does it change ??

Answer 26

According to me the answer is none of above are correct.

Answer 27

Because capacitance becomes half
Potential becomes twice

Answer 28

Charge is constant

Answer 29

electric field remains constant

Answer 30

my initial guess was ... the potential remains constant.

as the capacitor is charged, there is no current ... hence the pd must be equal to emf.