The charge on the square plates of a parallel-plate capacitor is q. the potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. the amount of charge on the plates is now equal toa)4qb)2qc)qd)q/2e)q/4

D) Q/2

Explanation:

The relationship between charge Q, capacitance C and voltage drop V across a capacitor is

(1)

In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by

where is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.

Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: . The capacitance becomes

And therefore, the new charge stored on the capacitor will be

this means that the temperature of both substances will be equal

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m