The potential difference between the two plates of a parallel plate condenser is $250 v o l t$ and the distance between them is $5 c m$ . The uniform electric field intensity is:

5000 V m^{- 1}

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

50 V m^{- 1}

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

500 V m^{- 1}

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

250 V m^{- 1}

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

Open in App

Solution

The correct option is B $5000 V m^{- 1}$
We know $E = \frac{V}{d}$
$\Rightarrow E = \frac{250}{5 \times 10^{- 2}}$ $= 5000 V / m$

Suggest Corrections

Similar questions

Q. The potential difference between two parallel plates 1 cm apart is 100V. The electric field strength between them is :

Two parallel metal plates are $3 m m$ apart and have a uniform electric field strength $500$ $V m^{- 1}$ between them. The potential difference between the plates is:

Two parallel plates are 0.03m apart. The electric field intensity between them is 3000 newton per coulomb. Calculate the potential difference between the plates?

Q. The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :

Q. A

1 m m

thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

Join BYJU'S Learning Program

The potential difference between the two plates of a parallel plate condenser is 250 volt and the distance between them is 5 cm. The uniform electric field intensity is:

The correct option is B 5000 Vm−1We know E=Vd⇒E=2505×10−2=5000 V/m

The potential difference between the two plates of a parallel plate condenser is $250 v o l t$ and the distance between them is $5 c m$ . The uniform electric field intensity is:

The correct option is B $5000 V m^{- 1}$
We know $E = \frac{V}{d}$
$\Rightarrow E = \frac{250}{5 \times 10^{- 2}}$ $= 5000 V / m$