Consider a parallel plate capacitor, of ...

Consider a parallel plate capacitor, of capacitance C, plate area A and plate separation d. It is being charged by using a battery. The quantity `epsilon_(0)(d phi_(E))/(dt)` has the dimensions of [`epsilon_(0)`=permittivity of free space, `phi_(E)` is the electric flux, dt = time]

e.m.f.

current

resistance

frequency

Text Solution

Verified by Experts

The correct Answer is:

`epsilon_(0) (d phi_(E))/(dt) = epsilon_(0)(d)/(dt) [EA] [ :' phi_(E)` = Electric field `xx` Area]
`=epsilon_(0)d/(dt)[V/d cdot A] = (epsilon_(0)A)/(d) ((dV)/(dt))`
but `(epsilon_(0)A)/(dt) = C` = Capacity of the capacitor
`= C(dV)/(dt)`
`:. epsilon_(0) (dphi_(E))/(dt) = [Q/V xx V/t] [ :' C = Q/V]`
`:. epsilon_(0)(dphi_(E))/(dt) = [Q/T] = 1 [ :' I = Q/t]`
Thus it has the dimension of current.

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The dimensions of 1/2 epsilon_(0)E^(2) (epsilon_(0)= permittivity of free space, E= electric field) is

`[MLT^(-1)]`

`[ML^(-1)T^(-2)]`

`[MLT^(-2)]`

`[ML^(2)T^(-1)]`

The dimension of ((1)/(2))epsilon_(0)E^(2) ( epsilon_(0) : permittivity of free space, E electric field

(a) `MLT^(-1)`

(b) `ML^(2)T^(2)`

( c ) `ML^(-1)T^(-2)`

( d ) `ML^(2)T^(-1)`

Find the dimension of epsilon_(0)E^(2) , where epsilon_(0) is permittivity of free space and E is the electricity field

`[M^(1) L^(1) T^(-2) A^(1)]`

`[M^(1) L^(0)T^(-1)]`

`[M^(1) L^(-1) T^(-2) A^(1)]`

`[M^(1) L^(-1) T^(-2)]`