A parallel-plate capacitor has plates of area 200 cm 2 and separation between the plates 1 mm- The potential difference developed if a charge of 1 nC i-e-1 - 10-9 C is given to the capacitor is something-then if the plate separation is now increased to 2 mm- what will be the new potential difference-

We know that Q/C=V nbsp; nbsp; nbsp; nbsp; nbsp; .......(i) And also C=ε_0 A/d Putting this in eq(i) V= 1 × 10^ 9×1×10^ 3/200×10^ 4×8.85×10^ 12=5. ...

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

Byju's Answer

Standard XII

Physics

Potential and Potential Difference

A parallel-pl...

Question

A parallel-plate capacitor has plates of area $200 c m^{2}$ and separation between the plates $1 m m$ . The potential difference developed if a charge of $1 n C (i . e . 1 \times 10^{- 9} C)$ is given to the capacitor is something,then if the plate separation is now increased to $2 m m$ , what will be the new potential difference?

Open in App

Solution

We know that
$\frac{Q}{C} = V$ .......(i)
And also $C = \frac{ε_{0} A}{d}$
Putting this in eq(i)
$V = \frac{1 \times 10^{- 9} \times 1 \times 10^{- 3}}{200 \times 10^{- 4} \times 8.85 \times 10^{- 12}} = 5.65 V$
If plate separation $(d)$ is made $2 mm$ then potential difference $(V^{'})$ is given by
$V^{'} = \frac{Q}{C} = \frac{1 \times 10^{- 9} \times 2 \times 10^{- 3}}{200 \times 10^{- 4} \times 8.85 \times 10^{- 12}} = 11.3 V$

Suggest Corrections

Similar questions

Q. A parallel-plate capacitor has plates of area

A

and separation

d

and is charged to a potential difference

V

. The charging battery is then disconnected and the plates are pulled apart until their separation is

2 d

. What is the work required to separate the plates?

Q. The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?
(a) Charge on the capacitor
(b) Potential difference across the capacitor
(c) Energy of the capacitor
(d) Energy density between the plates

Q. A parallel plate air capacitor has capacity

^{'} C^{'}

, distance of separation between plate is

^{'} d^{'}

and potential difference

^{'} V^{'}

is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

Q. The plates of a parallel plate capacitor are charged upto

100 V

. now, after removing the battery, a

2 m m

thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by

1.6 m m

. The dielectric constant of the plate is :

A parallel plate capacitor has rectangular plates each of area $20 c m^{2}$ is separated by a distance D the potential difference between the two plates?

Join BYJU'S Learning Program

We know that QC=V .......(i) And also C=ε0Ad Putting this in eq(i) V=1×10−9×1×10−3200×10−4×8.85×10−12=5.65 V If plate separation(d) is made 2 mm then potential difference (V′) is given by V′=QC=1×10−9×2×10−3200×10−4×8.85×10−12=11.3 V