A spring with an unstrained length of 0.074 m and a spring constant of 2.4 N/m h

Question

A spring with an unstrained length of 0.074 m and a spring constant of 2.4 N/m hangs vertically downward from the ceiling. A uniform electric field directed upward fills the region containing the spring. A sphere with a mass of 5.9 x 10-3 kg and a net charge of +7.1 µC is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is 0.064 m. What is the magnitude of the external electric field?

Helpis much appreciated thanks!

Explanation / Answer

   Change in length of spring   x   =   0.074   -   0.064                                                          =   0.010   m    Force on spring   F   =   k * x                                     =   2.4 * 0.010                                     =   0.024   N    Net force on the spring is the difference of Coulomb's force and weight, i.e.    FC - m * g   =   F    FC -   5.9 * 10-3 * 9.8   =   0.024    FC   =   0.024 +   5.782 * 10-2   N             =   8.182 * 10-2   N    Since   FC =   E * q    hence electric field strength      E   =   8.182 * 10-2 / 7.1 * 10-6                                                             =   1.152 * 104   N/C

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