A 73.0-g block carrying a charge Q = 33.0 Mu C is connected to a spring for whic

Question

A 73.0-g block carrying a charge Q = 33.0 Mu C is connected to a spring for which k = 70.0 N/m. The block lies on a frictionless, horizontal surface and is immersed in a uniform electric field of magnitude E = 4.80 X 10^4 N/C directed as shown in the figure below. The block is released from rest when the spring is unstretched (x = 0). (a) By what maximum distance does the block move from its initial position? cm (b) Find the subsequent equilibrium position of the block and the amplitude of its motion. position cm Amplitude cm (C) Using conservation of energy, find a symbolic relationship giving the potential difference between its initial position and the point of maximum extension in terms of the spring constant k, the amplitude A, and the charge Q. triangle V =

Explanation / Answer

a)

let the distance is d

now, for the maximum distance ,

0.5 * k * d^2 = E * q * d

0.5 * 70 * d^2 = 4.80 *10^4 * 33 *10^-6 * d

solving for d

d = 0.0452 m

the maximum distance the block move is 0.0452 m

b)

for the equilibrium position ,

k*x = q*E

70 * x = 33 *10^-6 * 4.80 *10^4

solving for x

x = 0.0226 m

x = 2.26 cm

the position is 2.26 cm

amplitude of the motion is 2.26 cm

c)

Potential difference = q * E * d

Potential difference = 33 *10^-6 * 4.8 *10^6 * 0.0452

Potential difference = 7.17 V

the Potential difference is 7.17 v

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