Physics, 31.07.2019 11:30 Casmereee

Aresearch van de graaff generator has a 2.00-mdiameter metal sphere with a charge of 5.00 mc on it. (a) what is the potential near its surface? (b) at what distance from its center is the potential 1.00 mv? (c) an oxygen atom with three missing electrons is released near the van de graaff generator. what is its energy in mev when the atom is at the distance found in part b?

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Answer from: tdyson3p6xvtu

Part A: The potential on the surface of the charged sphere is $4.5 \times 10^7$ volts.

Part B: The distance is $4.5 \times 10^{10}$ 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 $x = 4.5 \times 10^{10} \;\rm m$ 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.

$V = k\dfrac {Q}{R}$

Where V is the potential on the surface, Q is the charge, R is the radius and k is the Coulomb's constant.

$V = 8.99 \times 10^9\times \dfrac {5 \times 10^{-3}}{\dfrac {2}{2}}$

$V = 4.5 \times 10^7 \;\rm Volts$

Hence the potential on the surface of the charged sphere is $4.5 \times 10^7$ volts.

Part B

The potential of 1 mv generated at a certain distance x from the centre of the sphere is given below.

$V_x = k \dfrac {Q}{x}$

$1 \times 10^{-3} = 8.99 \times 10^9 \times \dfrac{5 \times 10^{-3}} {x}$

$x = 4.5 \times 10^{10} \;\rm m$

Hence the distance is $4.5 \times 10^{10}$ 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.

U = QV

At the distance found in part b, which is $x = 4.5 \times 10^{10} \;\rm m$ , the energy of the oxygen atom is,

$U = +3e \times 1 \times 10^{-3}$

$U = 3me \;\rm V$

Hence the energy of the oxygen atom found at the distance of $x = 4.5 \times 10^{10} \;\rm m$ is 3me V.

To know more about potential, follow the link given below.

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Answer from: cardsqueen

(a) The potential on the surface of a charged sphere of radius R is equal to
$V(R) = k_e \frac{Q}{R}$
where $k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant,

is the charge on the sphere's surface.
For the generator mentioned in the problem, the charge is $Q= 5 mC=5 \cdot 10^{-3} C$ , while the radius is $R= \frac{d}{2}= \frac{2.0 m}{2} =1.0 m$ . Using these values in the formula, we can calculate the potential at the surface:
$V(R)=8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3} C}{1.0 m}=4.5 \cdot 10^7 V$

(b) The potential generated by the sphere at a certain distance r from the centre of the sphere is given by
$V(r) = k_e \frac{Q}{r}$
the problem asks at which distance $V(r) = 1 mV=1\cdot 10^{-3} V$ . Substituting in the previous formula we can find the value of r:
$r=k_e \frac{Q}{V(r)}= 8.99 \cdot 10^9 N m^2 C^{-2} \frac{5 \cdot 10^{-3}}{1\cdot 10^{-3} V}=4.5 \cdot 10^{10} m$

(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
U=q V

where q is the charge of the particle, which is in our case q=+3e

. So we can calculate the energy of the oxygen atom at the distance found in part b, which corresponds to $r=4.5 \cdot 10^{10}m$ and a voltage of V=1 mV