Respuesta :
(a) The potential on the surface of a charged sphere of radius R is equal to
[tex]V(R) = k_e \frac{Q}{R} [/tex]
where [tex]k_e = 8.99 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant, [tex]Q [/tex] is the charge on the sphere's surface.
For the generator mentioned in the problem, the charge is [tex]Q= 5 mC=5 \cdot 10^{-3} C[/tex], while the radius is [tex]R= \frac{d}{2}= \frac{2.0 m}{2} =1.0 m [/tex]. Using these values in the formula, we can calculate the potential at the surface:
[tex]V(R)=8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3} C}{1.0 m}=4.5 \cdot 10^7 V [/tex]
(b) The potential generated by the sphere at a certain distance r from the centre of the sphere is given by
[tex]V(r) = k_e \frac{Q}{r} [/tex]
the problem asks at which distance [tex]V(r) = 1 mV=1\cdot 10^{-3} V[/tex]. Substituting in the previous formula we can find the value of r:
[tex]r=k_e \frac{Q}{V(r)}= 8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3}}{1\cdot 10^{-3} V}=4.5 \cdot 10^{10} m[/tex]
(c) An oxygen atom with 3 missing electrons has a positive charge of +3e, with e being the elementary charge.
The electric potential energy of a charged particle located at some point with voltage V is
[tex]U=q V[/tex]
where q is the charge of the particle, which is in our case [tex]q=+3e[/tex]. So we can calculate the energy of the oxygen atom at the distance found in part b, which corresponds to [tex]r=4.5 \cdot 10^{10}m[/tex] and a voltage of [tex]V=1 mV[/tex]:
[tex]U=(3 e)(1 mV) = 3 meV[/tex]
[tex]V(R) = k_e \frac{Q}{R} [/tex]
where [tex]k_e = 8.99 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant, [tex]Q [/tex] is the charge on the sphere's surface.
For the generator mentioned in the problem, the charge is [tex]Q= 5 mC=5 \cdot 10^{-3} C[/tex], while the radius is [tex]R= \frac{d}{2}= \frac{2.0 m}{2} =1.0 m [/tex]. Using these values in the formula, we can calculate the potential at the surface:
[tex]V(R)=8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3} C}{1.0 m}=4.5 \cdot 10^7 V [/tex]
(b) The potential generated by the sphere at a certain distance r from the centre of the sphere is given by
[tex]V(r) = k_e \frac{Q}{r} [/tex]
the problem asks at which distance [tex]V(r) = 1 mV=1\cdot 10^{-3} V[/tex]. Substituting in the previous formula we can find the value of r:
[tex]r=k_e \frac{Q}{V(r)}= 8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3}}{1\cdot 10^{-3} V}=4.5 \cdot 10^{10} m[/tex]
(c) An oxygen atom with 3 missing electrons has a positive charge of +3e, with e being the elementary charge.
The electric potential energy of a charged particle located at some point with voltage V is
[tex]U=q V[/tex]
where q is the charge of the particle, which is in our case [tex]q=+3e[/tex]. So we can calculate the energy of the oxygen atom at the distance found in part b, which corresponds to [tex]r=4.5 \cdot 10^{10}m[/tex] and a voltage of [tex]V=1 mV[/tex]:
[tex]U=(3 e)(1 mV) = 3 meV[/tex]
Part A: The potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.
Part B: The distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.
Part C: The energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.
How do you calculate the potential?
Given that the diameter of the sphere is 2.00 m and the charge of 5.00 mc.
Part A
The potential on the surface of a charged sphere is given below.
[tex]V = k\dfrac {Q}{R}[/tex]
Where V is the potential on the surface, Q is the charge, R is the radius and k is the Coulomb's constant.
[tex]V = 8.99 \times 10^9\times \dfrac {5 \times 10^{-3}}{\dfrac {2}{2}}[/tex]
[tex]V = 4.5 \times 10^7 \;\rm Volts[/tex]
Hence the potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.
Part B
The potential of 1 mv generated at a certain distance x from the centre of the sphere is given below.
[tex]V_x = k \dfrac {Q}{x}[/tex]
[tex]1 \times 10^{-3} = 8.99 \times 10^9 \times \dfrac{5 \times 10^{-3}} {x}[/tex]
[tex]x = 4.5 \times 10^{10} \;\rm m[/tex]
Hence the distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.
Part C
Given that an oxygen atom with three missing electrons is released near the van de-Graaff generator. It means that the charge Q = +3e
The electric potential energy of a charged particle located at some point with voltage V is given below.
[tex]U = QV[/tex]
At the distance found in part b, which is [tex]x = 4.5 \times 10^{10} \;\rm m[/tex], the energy of the oxygen atom is,
[tex]U = +3e \times 1 \times 10^{-3}[/tex]
[tex]U = 3me \;\rm V[/tex]
Hence the energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.
To know more about potential, follow the link given below.
https://brainly.com/question/1313684.
