Two positive charges 12×10^-6C and 8×10^-6 are 10cm apart. Find the workdone in bringing them 4cm together

W = 5.76 J

Explanation:

Given that,

Charge 1,

Charge 2,

The distance between charges is 10 cm.

Initial distance = 10 cm = 0.1 m

Final distance = 10 cm - 4 cm = 6 cm = 0.06 m

We need to find the work done in bringing the 4 cm together.

We know that,

Work done = change in kinetic energy

So, the required work done is 5.76 J.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j