Electric force and coulomb balance (need solution for 4.-5.-6.) The figure opposite shows the details of the Coulomb balance. It essentially consists of a flat and rigid frame, permanently mounted on one of the ends of the device. A second identical armature mounted on a pivot can be taken at different distances from the first armature. The gravitational forces which are exerted on the second armature are balanced by a counterweight of modifiable position. The oscillations of the mobile part are quickly damped by a set of magnets which operates by the effect of eddy currents in the adjoining vertical conductive plate. A large potential difference is used to charge the two armatures. The upper, mobile, is in a position of equilibrium determined by the electric force (Fe), which is a downward and gravitational attraction: upward forces attributable to the counterweights (Fcp) + downward force attributable to the weight of the plate mobile (Fpm). The displacements d mobile are indeed very small, and to appreciate these displacements in a precise manner, a comparator microscope is used. 1. By considering this assembly as being infinite plane plates uniformly loaded with the same load in absolute values, but of opposite signs; obtain a relation giving the electric field (Etot) between the plates as a function of the potential difference (delta V) between them and the distance (d) which separates the plates. 2. Write the (simple) expression of the electric force on the movable plate having a charge Qmobile immersed in the electric field generated by the fixed plate (Efixe), see image on the left. 3. From the notions of uniformly charged infinite flat plates, determine a relation giving the value of the charge of the movable plate (Qmobile) from the electric field (Emobile) that it generates. 4. Substitute the relation obtained in #3 into the electric force relation from #2 and use the electric field relation from #1 to obtain an equation of electric force as a function of potential difference (delta V), knowing that Emovable = EFixed. 5. This electric force on the movable plate will be equal in intensity to the weight (FgMd) of a mass deposited on the movable plate, image on the right. Write this equality and replace the weight by a simple relation according to the mass and the electric force by the relation obtained in #4. 6. From the equation of #5, establish an equation giving the balancing potential difference at the terminals of the plates as a function of the deposited mass.
Electric force and coulomb balance (need solution for 4.-5.-6.)
The figure opposite shows the details of the Coulomb balance. It essentially consists of a flat and rigid frame, permanently mounted on one of the ends of the device. A second identical armature mounted on a pivot can be taken at different distances from the first armature. The gravitational forces which are exerted on the second armature are balanced by a counterweight of modifiable position. The oscillations of the mobile part are quickly damped by a set of magnets which operates by the effect of eddy currents in the adjoining vertical conductive plate. A large potential difference is used to charge the two armatures. The upper, mobile, is in a position of equilibrium determined by the electric force (Fe), which is a downward and gravitational attraction: upward forces attributable to the counterweights (Fcp) + downward force attributable to the weight of the plate mobile (Fpm). The displacements d mobile are indeed very small, and to appreciate these displacements in a precise manner, a comparator microscope is used.
1. By considering this assembly as being infinite plane plates uniformly loaded with the same load in absolute values, but of opposite signs; obtain a relation giving the electric field (Etot) between the plates as a function of the potential difference (delta V) between them and the distance (d) which separates the plates.
2. Write the (simple) expression of the electric force on the movable plate having a charge Qmobile immersed in the electric field generated by the fixed plate (Efixe), see image on the left.
3. From the notions of uniformly charged infinite flat plates, determine a relation giving the value of the charge of the movable plate (Qmobile) from the electric field (Emobile) that it generates.
4. Substitute the relation obtained in #3 into the electric force relation from #2 and use the electric field relation from #1 to obtain an equation of electric force as a function of potential difference (delta V), knowing that
Emovable = EFixed.
5. This electric force on the movable plate will be equal in intensity to the weight (FgMd) of a mass deposited on the movable plate, image on the right. Write this equality and replace the weight by a simple relation according to the mass and the electric force by the relation obtained in #4.
6. From the equation of #5, establish an equation giving the balancing potential difference at the terminals of the plates as a function of the deposited mass.
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