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A Van de Graaff generator is charged so that a proton at its surface accelerates radially outward at 1.35 ✕ 1012 m/s2. Find the following. (a) the magnitude of the electric force on the proton at that instant magnitude N direction ---Select--- (b) the magnitude and direction of the electric field at the surface of the generator magnitude N/C direction ---Select---

Respuesta :

hopemichael95

Answer:

(a).

We know that force is

F = m a

So

F = (1.67 x 10^(-27) x (1.38 x10^12)

F = 2.3 x 10^-15 N facing the radially outward direction

(b).

Similarly Force for charge is

F = q E = m a

So relating

E = F/q = 2.3x 10^-15 /(1.6 10^ -19

E = 144.75 N/C facing the radially outward direction

pristina

The answer to the question is

(a) The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.

(b) The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.

The answer can be explained as shown below.

Given that a proton accelerates radially outward at [tex]1.35\times 10^{12}\,m/s^2[/tex].

  • ie; [tex]a=1.35\times 10^{12}\,m/s^2[/tex]
  • We know the mass of a proton, [tex]m_p = 1.67\times10^{-27}\,kg[/tex].

From Newton's second law we have,

  • [tex]F=mg[/tex]

But here, [tex]m=m_p[/tex]

So, the electric force on the proton is;

  • [tex]F = m_p\, a= (1.67\times10^{-27}\,kg)\, \times(1.35\times 10^{12}\, m/s^2)=2.254\times10^{-15}\,N[/tex]
  • The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.

Also, we know that, in electrostatics,

  • [tex]F=Eq\\ \\\implies E=\frac{F}{q}[/tex]
  • The charge of a proton is, [tex]q=1.6\times 10^{-19}\,C[/tex]

Therefore, the electric field is given by,

  • [tex]E=\frac{2.254\times 10^{-15}N}{1.6\times 10^{-19}\,C} = 14031.25\, N/C[/tex]
  • The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.

Learn more about the electric force here:

https://brainly.com/question/8960054

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