A 12-V battery is connected to two metallic plates, creating a uniform electric

Question

A 12-V battery is connected to two metallic plates, creating a uniform electric field between them.

a) A -6.0 nC charge is released between the plates. Towards which plate will it move?
Explain.

b) The separation between the plates is 2.00 cm. Calculate the magnitude of the uniform electric field between them.

c) A positive charge q is released somewhere in between the two plates, it then travels a distance of 1.00 cm. Calculate the potential difference = Vf - Vi between the final and initial positions.

d) Calculate the final velocity of the moving charge at the end of its 1.00 cm
displacement if q = +5.0µC and its mass is m = 1.2 x10^-5 kg

Explanation / Answer

a. it will move toward the positive plate because it is negatively charged.

b. magnitude of E is (12 V)/(0.02 m) = 600 V/m

c. -6V becuase it is moving 0.01 m in a 600 V/m field, and negative because a positive charge will move from positive to negative so the final potential is lower than the initial.

d. W = qV = (5.0 * 10-6 C)(6 V) = 3.0 * 10-5 J       work done on the particle, which now must be its KE

(1/2)mv2 = 3.0 * 10-5 J

v = ((6.0 * 10-5 J/1.2 * 10-5 kg))1/2 = 2.236 m/s

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