# 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).

#### Solution

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]

=-144xx10^-3xx(-25/3)

=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
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 12 | Page 88