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a. If 5 J of work is done in moving 0.25 C of positive charge from point A to point B, what isthe difference in potential between the points?b. How much velocity change will a xenon ion achieve in moving across a potential differenceof 500 V if this potential drop occurs over Case 1) Imm, or Case 2) 1cm?c. A spherical shell of a Van de Graaf generator is to be charged to a potential of 1 millionVolts. Calculate the minimum shell radius if the dielectric strength of air is 3x106 V/m.

User Gavi
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Answer:a. To find the potential difference between points A and B, we need to use the formula:

ΔV = W / q

where ΔV is the potential difference, W is the work done, and q is the charge.

Plugging in the given values, we get:

ΔV = 5 J / 0.25 C = 20 V

Therefore, the potential difference between points A and B is 20 volts.

b. The kinetic energy gained by a charged particle as it moves through a potential difference is given by:

KE = qΔV

where KE is the kinetic energy, q is the charge, and ΔV is the potential difference.

Assuming that the xenon ion is a singly ionized ion with a charge of +1.6 × 10^-19 C, we can calculate the kinetic energy gained in each case as:

Case 1: ΔV = 500 V, d = 0 (immediate)

KE = qΔV = (1.6 × 10^-19 C) × (500 V) = 8 × 10^-17 J

Case 2: ΔV = 500 V, d = 1 cm

The electric field between the two points is given by:

E = ΔV / d = 500 V / (0.01 m) = 5 × 10^4 V/m

The force on the ion is given by:

F = qE = (1.6 × 10^-19 C) × (5 × 10^4 V/m) = 8 × 10^-15 N

Using the work-energy theorem, we can calculate the distance traveled by the ion as:

W = Fd = KE

d = KE / F = (8 × 10^-17 J) / (8 × 10^-15 N) = 0.01 m

Therefore, the ion travels a distance of 1 cm before it comes to rest.

c. The electric field at the surface of the spherical shell is given by:

E = V / r

where V is the potential and r is the radius of the shell.

The dielectric strength of air is the maximum electric field that air can withstand before it breaks down and becomes conductive. In this case, the dielectric strength of air is 3 × 10^6 V/m.

Therefore, we can write:

E = 3 × 10^6 V/m

V / r = 3 × 10^6 V/m

Solving for r, we get:

r = V / E = (1 × 10^6 V) / (3 × 10^6 V/m) = 0.333 m

Therefore, the minimum radius of the spherical shell is 0.333 meters or 33.3 cm.

User Sandeep Giri
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