Respuesta :
Answer:
Electric field, [tex]E=1.16\times 10^7\ N/C[/tex]
Explanation:
Charge in van de Graff generator, [tex]q=8.1\ mC=8.1\times 10^{-3}\ C[/tex]
Distance from center of generator, d = 2.5 m
The electric field at a distance d from the center of any charged particle is given by :
[tex]E=\dfrac{kq}{d^2}[/tex]
[tex]E=\dfrac{9\times 10^9\times 8.1\times 10^{-3}}{(2.5)^2}[/tex]
[tex]E=1.16\times 10^7\ N/C[/tex]
So, the the electric field 2.5 m from the center of the terminal of a Van de Graaff is [tex]E=1.16\times 10^7\ N/C[/tex]. Hence, this is the required solution.
Answer:
The electric field is [tex]116.6\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Distance= 2.5 m
Charge = 8.10 mC
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{kQ}{r^2}[/tex]
Where, k = Boltzmann constant
Q = charge
r = distance
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times8.10\times10^{-3}}{(2.5)^2}[/tex]
[tex]E=11664000\ N/C[/tex]
[tex]E=116.6\times10^{5}\ N/C[/tex]
Hence, The electric field is [tex]116.6\times10^{5}\ N/C[/tex]
