Question

Two positive charges of 12 µC and 8 µC respectively are 10 cm apart. Find the work done in bringing them 4 cm closer, so that, they are 6 cm apart.

Answer 1

Given conditions ⇒

q₁ = 12 μC = 12 × 10⁻⁶ C.
q₂ = 8 μC = 8 × 10⁻⁶ C.
Distance between them(r₁) = 10 cm. = 0.1 m.
Final Distance between them(r₂) = 6 cm . = 0.06 m.

Work done in bringing the charges upto the distance of 6 cm is given by,

W = k q₁ q₂ (1/r₂ - 1/r₁) 
= 9 × 10⁹ × 12 × 10⁻⁶ × 8 × 10⁻⁶(1/0.06 - 1/0.1)
= 864 × 10⁻³ (16.67 - 10)
= 864 × 10⁻³(6.67)
= 5760 × 10⁻³
= 5.76 J.

Hence, the work done is 5.76 J.


Hope it helps.

Answer 2

Answer:

The work done will be 5.76 J.

Explanation:

Given that,

First charge$q_{1}=12\times10^{-6}\ C$

Second charge$q_{2}=8\times10^{-6}\ C$

The distance $r_{1}= 10\ cm=0.1\ m$

The distance $r_{2}= 6\ cm=0.6\ m$

The work done will be

$W= kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})$

$W=9\times10^{9}\times12\times10^{-6}\times8\times10^{-6}\times(\dfrac{1}{0.06}-\dfrac{1}{0.1})$

$W= 5.76\ J$

Hence, The work done will be 5.76 J.