How fast does the negatively charged body move in an electric field in the following case:
2 oppositely charged parallel plates with an electric field of 80 V/m have potential difference of 40 V. A tiny sphere of mass 50 mg with 2120 excess electrons is placed in the field and sits on the positive plate. It is then moved so it touches the negative place and released without a push. How fast is it going when it strikes the positive plate again?
1 Answer | Add Yours
The electrical field between the two plates is 80 V/m. The potential difference between them is 40 V. Using this, we can derive the distance between the plates which is 40/80 = 0.5 m.
The sphere has a mass of 50*10^-6 kg and a charge of 2120 electrons. The charge of each electron is -1.6 * 10^-19 C. 2120 electrons would give the sphere a charge of 2120*1.6*10^-19 C.
The electric field which is given in V/m can be converted to N/C.
So the field of 80 N/C exerts a force of 2120*1.6*10^-19*80 N on the sphere.
Force = mass*acceleration. The mass of the sphere is 50*10^-6, this gives its acceleration as 2120*1.6*10^-19*80/50*10^-6.
Starting from rest, the speed acquired in a distance of 0.5 m is sqrt(2*a*s), where a is the acceleration and s is the distance.
The velocity with which the sphere strikes the positive plate is : sqrt(2*0.5*2120*1.6*10^-19*80/(50*10^-6))
=> 2.2396 * 10^-5 m/s
The required velocity with which the sphere strikes the positive plate is 2.2396*10^-5 m/s.
We’ve answered 317,537 questions. We can answer yours, too.Ask a question