The displacement current of 4 . 425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10 6 V s - 1 . The area of each plate of the capacitor is 40 cm 2 . The distance between each plate of the capacitor is x × 10 - 3 m . The value of x is , (Permittivity of free space, ε 0 = 8 . 85 × 10 - 12 C 2 N - 1 m - 2 ) _______

Step-by-step solution

Displacement current is given by i d = ε 0 d ϕ E d t Or i d = ε 0 d d t E A , where E = q A ε 0 Or i d = ε 0 d d t q A A ε 0 = d q d t = d d t C V Or i d = C d V d t = ε 0 A d d V d t Putting the values, we have 4 . 425 × 10 - 6 = 8 . 85 × 10 - 12 × 40 × 10 - 4 × 10 6   d d = 2 × 10 - 6 × 10 - 4 × 10 6 × 40 d = 80 × 10 - 4 = 8 × 10 - 3   m Hence, value of x = 8 .