MyQuestionIcon

4

You visited us 4 times! Enjoying our articles? Unlock Full Access!

More about Q=CV

A charge of 2...

Question

A charge of $20 μ C$ is placed on the positive plate of an isolated parallel-plate capacitor of capacitance $10 μ F .$ Calculate the potential difference developed between the plates.

Open in App

Solution

Given that capacitance = $10 μ F,$

charge = $20 μ C$

By effective charge = $10 μ F$

$\Rightarrow$ Potential diff. = 1 V

Suggest Corrections

thumbs-up

3

Similar questions

+ 2 \cdot 0 \times 10^{- 8} C

is placed on the positive plate and a charge of

- 1 \cdot 0 \times 10^{- 8} C

on the negative plate of a parallel-plate capacitor of capacitance

1 \cdot 2 \times 10^{- 3} μ F

. Calculate the potential difference developed between the plates.

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. Calculate the capacitance (in farads) of a charged, parallel-plate capacitor. Given the potential difference between the plates is equal to

X

volts and the amount of charge on the POSITIVE plate is equal to

Y

Q. In an isolated parallel plate capacitor of capacitance

C

, the four surfaces have charges

Q_{1}, Q_{2}, Q_{3}

Q_{4}

as shown. The potential difference between the plates is:

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

The Idea of Capacitance

PHYSICS

Watch in App

explore_more_icon

Explore more

More about Q=CV

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon