Question

A capacitor of capacitance 'c' is charged to 'v' volts by a battery. After sometime, the battery is disconnected and the distance btw the plates is doubled. Now a slab of dielectric constant 1 (i)The electric field btw the plates of the capacitor.
(ii) The energy stored in each capacitor.
Justify your answer in each case.

Answer 1

Answer:

answer" (and any subsequent words) was ignored because we limit queries to 32 words.

Answer 2

Explanation:

: Without the dielectric and with standard <br> notaions, `C = (in_(0) A)/(d)` <br> When battery is not disconnected, potential difference <br> V = constant <br> d' = 3d and k = 10 <br> `C' = (K in_(0)A)/(d') = (10in_(0) A)/(3d) = (10)/(3) ((in_(0))/(d)) = (10)/(3) C` <br> Charge on capacitor <br> `Q' = C'V = (10)/(3) CV = (10)/(3) Q` <br> Energy density, <br> `rho = (1)/(2)K in_(0) (V^(2))/(d'^(2)) = (K)/(3^(2)) ((1)/(2) in_(0) (V^(2))/(d^(2))) = (10)/(9) rho`