You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Idea behind a PPC

One plate of ...

Question

One plate of a parallel plate capacitor is suspended from a beam of a physical balance as shown in figure. The area of each plate is $625 c m^{2}$ and the distance between these plates is $5 m m$ . If an additional mass $0.04 g m$ is placed in the other pan of the balance, then the potential difference required between the plates to keep it in equilibrium will be:

150 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

188 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

225 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

310 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $225 V$
One plate capacitor
parallel capacitor
The area of each plate $= 625 c m^{2}$
Distance between these plate $= 5 m m$
Additional mass $= 0.04 g m$
$C = \frac{E_{0} A}{d} = \frac{8.854 \times (625 \times 10^{- 4})}{5 \times 10^{- 3}} = 110.675 V = Q / C = \frac{0.04 \times 10^{- 3}}{110.675} = 225 V$

Suggest Corrections

Similar questions

Q. In Millikan’s oil drop experiment, an oil drop of radius

r

and charge

Q

is held in equilibrium between the plates of a charged parallel plate capacitor when the potential difference is

V

. To keep a drop of radius

2 r

and charge

2 Q

in equilibrium between the plates ,the potential difference required will be

Q. Figure shows a parallel plate capacitor with plate area

A

and plate separation

d

. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant

K

is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

Q. Two conducting plates of parallel plate capacitor are placed at a small distance from each other as shown in figure. a, b, c and d are charges on different surfaces.

Then, potential difference between plates (

V_{H} - V_{L}

) is

Q. A small sphere of mass

20 g

and having charge

0.5 m C

is suspended by a thread between the plates of a parallel plate capacitor as shown in the figure. What potential difference ( in volt) between the plates (separation

20 c m

) should be applied so that the thread makes an angle of

45^{0}

with the vertical?

In a parallel plate capacitor the potential difference of 150 V is maintained between the plates. If distance between the plates is 6 mm, what will be the electric field at points A and B?

Join BYJU'S Learning Program

The correct option is C 225 VOne plate capacitorparallel capacitorThe area of each plate=625cm2Distance between these plate=5mmAdditional mass=0.04gmC=E0Ad=8.854×(625×10−4)5×10−3=110.675V=Q/C=0.04×10−3110.675=225V