The work done in carrying a charge of $5 μ C$ from a point A to a point B in an electric field is 10mJ. The potential difference $(V_{B} - V_{A})$ is

+ 2kV

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

- 2 kV

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

+ 200 V

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

- 200 V

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

Open in App

Solution

The correct option is A + 2kV
Work done $W = Q (V_{B} - V_{A}) \Rightarrow (V_{B} - V_{A}) = \frac{W}{Q} = \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}} J / C = 2 k V$

Suggest Corrections

Similar questions

12 J of work has to be done against an existing electric field to take a charge of 0.01 from A to B. Hwo much is the potential difference $V_{B} - V_{A}$ ?

Q. The work done by electric field in carrying a point charge

5 μ C

from a point

A

to point

B

10 mJ

. The potential difference

V_{B} - V_{A}

is:

Electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other.

When the particle of charge 10MC is brought from infinite to a point in electric field 10mJ work is done by external forces. What is the potential at that time.....??

Q. In figure two points

A

and

B

are located in a region of electric field. The potential difference

V_{B} - V_{A}

Join BYJU'S Learning Program

The work done in carrying a charge of 5μC from a point A to a point B in an electric field is 10mJ. The potential difference (VB−VA) is

The correct option is A + 2kV Work done W=Q(VB−VA)⇒(VB−VA)=WQ=10×10−35×10−6J/C=2kV

The work done in carrying a charge of $5 μ C$ from a point A to a point B in an electric field is 10mJ. The potential difference $(V_{B} - V_{A})$ is

The correct option is A + 2kV
Work done $W = Q (V_{B} - V_{A}) \Rightarrow (V_{B} - V_{A}) = \frac{W}{Q} = \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}} J / C = 2 k V$