Question

a) The electric potential at a position located a distance of 21.9 mm from a positive point charge of 8.20×10^-9C and 13.9 mm from a second point charge is 1.26 kV. Calculate the value of the second charge.
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b) The potential difference between two parallel conducting plates in vacuum is 175 V. An alpha particle with mass of 6.50×10^-27 kg and charge of 3.20×10^-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 29.0 cm.
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c) A positive charge of 4.80 μC is fixed in place. From a distance of 4.00 cm a particle of mass 5.20 g and charge +3.80 μC is fired with an initial speed of 76.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away?
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d) A charge Q₁ = 1.23 μC is at rest and is located 2.10 cm away from another fixed charge Q₂ = 1.75 μC. The first charge is then released. Calculate the kinetic energy of charge Q₁ when it is 5.50 cm away from charge Q₂.
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e) To recharge a 15.0 V battery, a battery charger must move 3.00×10⁵ C of charge from the negative terminal to the positive terminal. How much work is done by the battery charger?
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Please help with this asap w/ work! Thanks!

Answer 1

5-

work done = change * potential difference

W = Q * V

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suppose (+dq) is removed from nuetral conductor (say -ve terminal), then it lost charge (so becmes >>> - dq, and acquires negative potential), and other got (+dq, got +ve potential) so that becomes +vely charged

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since the 2 terminals have now dq charge and (V) potential difference

small work done (dW) = V dq

total work done W = V Q = 15 * 3.0*10^5 = 4.5*10^7 Joules

= 4.5 Mega joules