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What is the potential 0.530 × 10 −10 m from a proton (the average distance between the proton and electron in a hydrogen atom)?

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(a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential 5.00 MV? (b) What does your answer imply about the practical aspect of isolating such a large charge?

a. V = k Q r r = 1.80 km ; b. A 1-C charge is a very large amount of charge; a sphere of 1.80 km is impractical.

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What are the sign and magnitude of a point charge that produces a potential of –2.00 V at a distance of 1.00 mm?

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In one of the classic nuclear physics experiments at the beginning of the twentieth century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?

The alpha particle approaches the gold nucleus until its original energy is converted to potential energy. 5.00 MeV = 8.00 × 10 −13 J , so
E 0 = q k Q r r = 4.54 × 10 −14 m
(Size of gold nucleus is about 7 × 10 −15 m ).

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Additional problems

A 12.0-V battery-operated bottle warmer heats 50.0 g of glass, 2.50 × 10 2 g of baby formula, and 2.00 × 10 2 g of aluminum from 20.0 °C to 90.0 °C . (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes 5.00 min to warm the formula? ( Hint: Assume that the specific heat of baby formula is about the same as the specific heat of water.)

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A battery-operated car uses a 12.0-V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 × 10 2 -m high hill, and finally cause it to travel at a constant 25.0 m/s while climbing with 5.00 × 10 2 -N force for an hour.

E tot = 4.67 × 10 7 J E tot = q V q = E tot V = 3.89 × 10 6 C

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(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

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A uniformly charged ring of radius 10 cm is placed on a nonconducting table. It is found that 3.0 cm above the center of the half-ring the potential is –3.0 V with respect to zero potential at infinity. How much charge is in the half-ring?

V P = k q tot z 2 + R 2 q tot = −3.5 × 10 −11 C

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A glass ring of radius 5.0 cm is painted with a charged paint such that the charge density around the ring varies continuously given by the following function of the polar angle θ , λ = ( 3.0 × 10 −6 C/m ) cos 2 θ . Find the potential at a point 15 cm above the center.

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A CD disk of radius ( R = 3.0 cm ) is sprayed with a charged paint so that the charge varies continually with radial distance r from the center in the following manner: σ = ( 6.0 C/m ) r / R .

Find the potential at a point 4 cm above the center.

V P = −2.2 GV

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(a) What is the final speed of an electron accelerated from rest through a voltage of 25.0 MV by a negatively charged Van de Graff terminal? (b) What is unreasonable about this result? (c) Which assumptions are responsible?

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A large metal plate is charged uniformly to a density of σ = 2.0 × 10 −9 C/m 2 . How far apart are the equipotential surfaces that represent a potential difference of 25 V?

Recall from the previous chapter that the electric field E P = σ 0 2 ε 0 is uniform throughout space, and that for uniform fields we have E = Δ V Δ z for the relation. Thus, we get σ 2 ε 0 = Δ V Δ z Δ z = 0.22 m for the distance between 25-V equipotentials.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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