a capacitor is connected to a battery. the force of attraction btn plates whn the separation is halved???
a remains same.
b becoms 8 tims
c bcms 4 times
d bcms 2 tims
1357Force does not change.. bcos the electric field does not change... (use gauss's law)
3V=Q/C.
V is maintained by the battery.
but Q decreases inorder to maintain V.
F=QE=QV/dist..
so the net expression will lead to the unchangd force.
1u mean to say tht as both q and distance changes net value of force is same??
1u mean to say tht as both q and distance changes net value of force is same??
106think like this Electric field due to any one plate on the other E = σ/2ε if the distance between the plates d << A (area of plate) which is the case in case of a parallel plate capacitor.
So, F = Qσ/2ε
which is independent of the separation between the plates
66If the plate area is A , the separation d, and the voltage of the battery V, then the charge (in magnitude) on each plate is
q=\epsilon_0\dfrac{AV}{d}
Field of any one plate is E=\dfrac{q}{2\epsilon_0}=\dfrac{V}{2d}
So the force on the other plate is
F=qE=\dfrac{\epsilon_0AV^2}{2d^2}
So the other factors being the same, the force between the plates
F\propto \dfrac{1}{d^2}
Accordingly, if the separation is halved, the force becomes 4 times. Hence, (c).
