Question

The amount of work required to increase the distance between -6μC and 4μC from 6 cm to 18 cm will be :

  (a) 1.8 J
  (b) 2.4 J
  (c) 1.8 μJ
  (d) 2.4 μJ​

Answer 1

We have to find the amount of work required to increase the distance between -6μC and 4μC from 6cm to 18 cm.

Work done is the negative of change of potential energy.

i.e., Work done = -[final potential energy - initial potential energy]

= initial potential energy - initial potential energy

= $\frac{KQ_1Q_2}{r_1}-\frac{KQ_1Q_2}{r_2}$

= $KQ_1Q_2\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$

here, Q₁ = 6μC , Q₂ = 4μC , r₁ = 6cm = 0.06 m and r₂ = 18cm = 0.18 m

= 9 × 10^9 × (6 × 10^-6) × (4 × 10^6) × $\left[\frac{1}{0.06}-\frac{1}{0.18}\right]$

= 216 × 10^-3 × 1/0.09

= 2.4 J

Therefore the work done required to increase the distance between -6μC and 4μC is 2.4 J. so the correct option is (b)

Answer 2

Explanation:

v= kq/r. ,v=w/qo

kq/r =w/qo

9×10⁹×6×4×10²/12×10¹²

= 1.8 j