Home
Class 12
PHYSICS
As per this diagram a point charge +q is...

As per this diagram a point charge `+q` is placed at the origin `O`. Work done in taking another point charge `-Q` from the point `A(0, a)` to another point `B(a, 0)` along the staight path `AB` is:

A

`(qQ)/(4piepsi_(0))((a-b)/(ab))`

B

`(qQ)/(4piepsi_(0))((b-a)/(ab))`

C

`(qQ)/(4piepsi_(0))((b)/a^(2)-(1)/(b))`

D

`(qQ)/(4piepsi_(0))((a)/(b^(2))-(1)/(b))`

Text Solution

Verified by Experts

The correct Answer is:
A
Potential at point A is `V_A=1/(4piepsilon_0) q/a`
Potential at point B is `V_B=1/(4piepsilon_0) q/b`
Work done in taking a charge Q from A to B is
`W=Q(V_B-V_A)=(Qq)/(4piepsilon_0)[1/b-1/a]=(Qq)/(4piepsilon_0)["a-b"/"ab"]`
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NCERT FINGERTIPS|Exercise Potential Due To An Electric Dipole|5 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NCERT FINGERTIPS|Exercise Potential Due To A System Of Charges|7 Videos

  • ELECTROSTATIC POTENTIAL AND CAPACITANCE

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

  • ELECTROMAGNETIC WAVES

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

  • MAGNETISM AND MATTER

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

In the firgure the charge Q is at the centre of the circle. Work done is maximum when another charge is taken from point P to

A charge +q is placed at the origin O of X- Y axes as shown in the figure . The work done in taking a charge Q from A to B along the straight line AB is

A point charge q is placed at the origin . How does the electric field due to the charge very with distance r from the origin ?

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2xx10^(-9)C from a point P (0,0,3 cm) to a point Q (0,4 cm, 0) via a point R (0,6 cm, 9cm).

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2xx10^(-9) from a point P(0,0,3cm) to a point Q (0,4cm,0) via a point R(0,6cm,9cm).

As shown in the figure a charge +2 C is situated at the origin O and another charge +5 C is on the x -axis at the point A . The later charge from the point A is then brought to a point Bon the y - axis . The work done is ("given"(1)/(4piepsilon_(0))=9xx10^(9))

A point charge of 0.01 mu C is placed at the origin. Calculate the electric field at another point charge of 0.05 mu C placed at (0 m, 5 m).