Hii
Can someone please solve this??
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as we know that capacity of a parallel plate capacitor is given by
C = ɛA/ d ----(1)
where
q = charge given to capacitor
V = potential difference
A = Area of plates
ɛ = epsilon
d = distance b/w two plates
Putting C = q/V in 1 eqn.
q/V = ɛA/d
==> V = qd/ɛA ----(2)eqn.
If a dielectric glass slab of dielectric constant K is placed b/w two plates the the capacitance is given by
C = K ɛA /d -----(3)eqn.
After placing the dielectric slab the potential becomes 1/8
So C = q/ (V/8) = 8q/V
Putting in 3 eqn.
8q/V = K ɛA/d
==> KV = 8qd/ɛA ---(4)
On dividing (4) by (2),we get
K = 8
(Sorry but I am not able to divide here, if you are unable to understand you can try to do it in you copy or ask me in comments. Please reply in comment whether you understand or not)
HOPE IT HELPS
Thanks for asking this question and improving my knowledge
Thanks Nishita from the depth of my heart.
C = ɛA/ d ----(1)
where
q = charge given to capacitor
V = potential difference
A = Area of plates
ɛ = epsilon
d = distance b/w two plates
Putting C = q/V in 1 eqn.
q/V = ɛA/d
==> V = qd/ɛA ----(2)eqn.
If a dielectric glass slab of dielectric constant K is placed b/w two plates the the capacitance is given by
C = K ɛA /d -----(3)eqn.
After placing the dielectric slab the potential becomes 1/8
So C = q/ (V/8) = 8q/V
Putting in 3 eqn.
8q/V = K ɛA/d
==> KV = 8qd/ɛA ---(4)
On dividing (4) by (2),we get
K = 8
(Sorry but I am not able to divide here, if you are unable to understand you can try to do it in you copy or ask me in comments. Please reply in comment whether you understand or not)
HOPE IT HELPS
Thanks for asking this question and improving my knowledge
Thanks Nishita from the depth of my heart.
Anonymous:
yeah
great
thanks !
im asking one more ques
no need to said thanks
we are friends
if u can solve pls help
and also you ahve also increased my knowledge
ya i will try
ask the question
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