Two plates are 2cm apart and a potential difference of 10V is applied between them. The electric field between the plates is

20 N/C

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

500 N/C

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

5 N/C

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

250 N/C

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

Open in App

Solution

The correct option is D 500 N/C
Electric field $| E | = \frac{d V}{d r} = \frac{10}{2 c m}$
$= \frac{10}{2 \times 10^{- 2} m} = 500 N / C$

Suggest Corrections

Similar questions

Q. Two plates are

20 c m

apart and the potential difference between them is

10 V

. The electric field between the plates is

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. Two plates are at potentials

- 10 V

and

+ 30 V

. If the separation between the plates is

2 c m

then the electric field between them will be

Join BYJU'S Learning Program

Two plates are 2cm apart and a potential difference of 10V is applied between them. The electric field between the plates is

The correct option is D 500 N/CElectric field |E|=dVdr=102cm=102×10−2m=500N/C

The correct option is D 500 N/C
Electric field $| E | = \frac{d V}{d r} = \frac{10}{2 c m}$
$= \frac{10}{2 \times 10^{- 2} m} = 500 N / C$