Advertisement Remove all ads

Calculate the Work Done in Taking a Small Charge - Physics

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Advertisement Remove all ads


Charge located at the origin, q = 8 mC= 8 × 10−3 C

Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = − 2 × 10−9 C

All the points are represented in the given figure.

Point P is at a distance, d1 = 3 cm, from the origin along z-axis.

Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.

Potential at point P, `V_1=q/(4piin_0xxd_1)`

Potential at point Q,`V_2=q/(4piin_0xxd_2)`

Work done (W) by the electrostatic force is independent of the path.

`therefore W=q_1 [V_2-V_1]`

`=q_1[q/(4piin_0d_2)-q/(4piin_0d_1)]` .......(1)

Where , `1/(4piin_0)=9xx10^9 Nm^2C6-2`

`therefore W=9xx10^9xx8xx10^-3xx(-2xx10^-9)[1/0.04-1/0.03]`


`=1.27 J`

Therefore, work done during the process is 1.27 J.

Concept: Potential Energy in an External Field - Potential Energy of a Dipole in an External Field
  Is there an error in this question or solution?
Advertisement Remove all ads


NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 12 | Page 88
Advertisement Remove all ads
Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×