Q. 123.7( 21 Votes )

# A charge of 8 m C is located at the origin. Calculate the work done in taking a small charge of –2 × 10^{–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, 9 cm).

Answer :

Given,

Charge at origin (O), q_{0} = 8 × 10^{-3} C

The charge (q_{1} = -2 Nc) is moving through the given points, P(0, 0, 3)cm, R(0, 6, 9)cm and Q(0, 4, 0)cm respectively.

The picture below represents above situation,

Since, Work is independent on the path followed we concentrate on point P and Q only.

Potential at any point is given by,

Where,

q = charge

d = distance from origin

ϵ_{0} = permittivity of space

and,

Thus, Potential at point P,

(∵ distance of point from origin = 3 cm = 0.03m)

Potential at point Q,

Work done, W = q_{1} × (Change in potential difference)

= 1.27 J

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A charge 'q' is moved from a point A above a dipole of dipole moment 'p' to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

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