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3 years ago

What information is needed to calculate acceleration?

Physics

1 answer:

aalyn [17]3 years ago

3 0

There is many method to calculate acceleration

(v-u)/2

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3 years ago

Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char

IgorC [24]

Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

$K=\frac{1}{2}mv^2$ = 800J (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

$F=ma=qE\\\\a=\frac{qE}{m}$ (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

$a=\frac{qV}{md}$

Next, you take into account the following formula for the calculation of the final speed of the charge:

$v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}$

Next, you replace this value of v in (1):

$K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV$ = 880J (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

$v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}$

And the kinetic energy becomes:

$K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV$ (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

$K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J$

hence, the kinetic energy of the charge incresases to 1200J

4 0

3 years ago