A parallel-plate capacitor has plates of area 200 cm2 and separation between the plates 1 mm- The potential difference developed if a charge of 1 nC i-e- 1 - 10-9C 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 XIII

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

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?

Q. A parallel plate capacitor has rectangular plates of area

400 {cm}^{2}

and separated by

2 mm

with air as a medium. What is the magnitude of charge on each plate if the applied potential difference is

200 Volts

?
(Take

ϵ_{o} = 8.85 \times 10^{- 12} F/m

)

Q. Two parallel plate capacitors have their plate areas

100 {cm}^{2}

and

500 {cm}^{2}

respectively. If they have the same charge and potential difference, and the distance between the plates of first capacitor is

0.5 mm,

then the distance between the plates of second capacitor is

A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of $2 \times 10^{6} m s^{- 1}$ moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?

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 becomes

2 d

. What is the work required to separate the plates?

[

1 Mark

]

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