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

Byju's Answer

Standard XII

Physics

More about Q=CV

The potential...

Question

The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is
[AFMC 2005]

2 F

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

4 F

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

0.5 F

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

0.25 F

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

Open in App

Solution

The correct option is A 2 F
Potential difference between the plates
= 10 - (-10)
= 20 V

Q = 40 C
Q = CV
$\Rightarrow C = \frac{Q}{V} = \frac{40}{20} = 2 F$
C = 2F

Suggest Corrections

Similar questions

Q. The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is
[AFMC 2005]

A parallel-plate capacitor has a capacitance of

4 μ F

. Initial charge on the capacitor is

20 μ C

. Which of the following changes would increase potential difference between its plates to

10 V

Q. A capacitor of capacitance

2 μ F

is charged with a

10 V

battery. Now battery is disconnected and a dielectric of dielectric constant

K = 4

is inserted between the plates. What is the potential difference between the plates of the capacitor?

Q. A paralle plate capacitor having capacitance

12 pF

is charged by a battery to a potential difference of

10 V

between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant

6.5

is slipped between the plates. The work done by the capacitor on the slab is

Five identical plates are arranged such that each pair is a capacitor of capacitance 2 mF.
The plates are connected to a source of emf 10 V as shown. The charge on the plate C is

Join BYJU'S Learning Program

The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is [AFMC 2005]

The correct option is A 2 FPotential difference between the plates = 10 - (-10) = 20 V Q = 40 C Q = CV ⇒C=QV=4020=2F C = 2F

The potentials of the two plates of capacitor are + 10V and -10 V. The charge on one of the plates is 40 C. The capacitance of the capacitor is
[AFMC 2005]

The correct option is A 2 F
Potential difference between the plates
= 10 - (-10)
= 20 V

Q = 40 C
Q = CV
$\Rightarrow C = \frac{Q}{V} = \frac{40}{20} = 2 F$
C = 2F