{"id":1652,"date":"2015-12-19T14:55:01","date_gmt":"2015-12-19T14:55:01","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1652"},"modified":"2024-11-12T12:42:11","modified_gmt":"2024-11-12T12:42:11","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-superficies-equipotenciais-trabalho-da-forca-eletrostatica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrostatica\/superficies-equipotenciais-trabalho-da-forca-eletrostatica\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-superficies-equipotenciais-trabalho-da-forca-eletrostatica\/","title":{"rendered":"Superf\u00edcies Equipotenciais \u2013 Resolu\u00e7\u00e3o EN"},"content":{"rendered":"

EQUIPOTENTIAL SURFACES \u2013 RESOLUTION<\/span><\/h1>\n
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\u00a0Commented resolution of entrance exam exercises on<\/span><\/b><\/b><\/p>\n

Equipotential Surfaces \u2013 Work of the Electrostatic Force<\/span><\/b><\/p>\n

01-<\/span><\/b>\u00a0I and II are true (see theory) and III is false because the electric field vector varies inversely with the square of the distance from the point to the charge \u2014\u00a0\u00a0<\/b>\u00a0R- C<\/b><\/span>\u00a0<\/b><\/b><\/b><\/p>\n

02-<\/span><\/b>\u00a0Observe the following expression \u2014 W=qU \u2014 U=W\/q \u2014\u00a0<\/b>\u00a0R- D<\/b><\/span>\u00a0<\/b><\/b><\/b><\/p>\n

03-<\/span><\/b>\u00a0All points on the trajectory AB are equidistant from the center of the charge and consequently constitute an equipotential surface \u2014 V\u00a0<\/b>A<\/b><\/sub>\u00a0= V\u00a0<\/b>B<\/b><\/sub>\u00a0\u00a0 \u2014 W=q.(V\u00a0<\/b>A<\/b><\/sub>\u00a0\u00a0\u2013 V\u00a0<\/b>B<\/b><\/sub>\u00a0)=q.0 \u2014 W=0 \u2014\u00a0\u00a0\u00a0<\/b>the work of the electric force in this displacement is zero.<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

04-<\/span><\/b>\u00a0The lines of force (solid lines) leave the positive charges and reach the negative ones and are perpendicular to the equipotential surfaces (dashed lines) \u2014\u00a0\u00a0<\/b>\u00a0R- E<\/b><\/span>\u00a0<\/b><\/b><\/b><\/p>\n

05-<\/span><\/b>\u00a0Along the same equipotential surface, the potential is always the same and the potential difference is zero and, consequently, the\u00a0<\/b>work is\u00a0<\/b>\u00a0also\u00a0\u00a0<\/b>zero.<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/p>\n

06-<\/span><\/b>\u00a0False \u2014 the potential decreases\u00a0<\/b>\uf020\u00a0<\/b>I.<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/p>\n

False \u2014\u00a0\u00a0<\/span><\/b>\uf020\u00a0<\/span><\/b>II.\u00a0\u00a0<\/span><\/b>the lines are perpendicular<\/b><\/span>\u00a0<\/b><\/b><\/p>\n

False \u2014\u00a0\u00a0<\/span><\/b>\uf020\u00a0<\/span><\/b>III. \u00a0\u00a0<\/span><\/b>the surfaces are flat and parallel<\/b><\/span>\u00a0<\/b><\/b><\/p>\n

R- And<\/span><\/b><\/p>\n

07-<\/span><\/b>\u00a0a) U=Ed \u2014 (100 \u2013 50)=5.10\u00a0<\/b>2<\/b><\/sup>\u00a0.d \u2014 d=50\/5.10\u00a0<\/b>2<\/b><\/sup>\u00a0\u00a0 \u2014\u00a0\u00a0\u00a0<\/b>d=1,0.10\u00a0<\/b>-1<\/b><\/sup>\u00a0m<\/b><\/span>\u00a0<\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/b><\/p>\n

b) W\u00a0<\/span><\/b>AB<\/span><\/b><\/sub>\u00a0=q(V\u00a0<\/span><\/b>A<\/span><\/b><\/sub>\u00a0\u00a0\u2013 V\u00a0<\/span><\/b>B<\/span><\/b><\/sub>\u00a0)=2.10\u00a0<\/span><\/b>-6<\/span><\/b><\/sup>\u00a0(100 \u2013 50) \u2014 W\u00a0<\/span><\/b>AB<\/span><\/b><\/sub>\u00a0=10\u00a0<\/span><\/b>2<\/span><\/b><\/sup>\u00a0.10\u00a0<\/span><\/b>-6<\/span><\/b><\/sup>\u00a0\u00a0 \u2014\u00a0\u00a0<\/span><\/b>W\u00a0<\/span><\/b>AB<\/span><\/b><\/sub>\u00a0=1.0.10\u00a0<\/span><\/b>-4<\/span><\/b><\/sup>\u00a0J \u00a0<\/span><\/b><\/p>\n

08-<\/span><\/b>\u00a0Since the initial (A) and final (D) positions are coincident, the work along the three trajectories is the same, since for conservative forces such as the electric force, the work is independent of the trajectory and depends only on the final and initial positions \u2014\u00a0\u00a0<\/b>\u00a0R- B<\/b><\/span>\u00a0\u00a0<\/b><\/b><\/b><\/p>\n

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