{"id":895,"date":"2023-05-27T09:38:11","date_gmt":"2023-05-27T12:38:11","guid":{"rendered":"https:\/\/proexatas.com\/?page_id=895"},"modified":"2025-02-25T18:31:03","modified_gmt":"2025-02-25T21:31:03","slug":"eletrostatica","status":"publish","type":"page","link":"https:\/\/proexatas.com\/index.php\/eletrostatica\/","title":{"rendered":"Eletrost\u00e1tica"},"content":{"rendered":"\n<h4 id=\"wp-block-themeisle-blocks-advanced-heading-47bbab7e\" class=\"wp-block-themeisle-blocks-advanced-heading wp-block-themeisle-blocks-advanced-heading-47bbab7e\"><em><strong>Lei de Coulomb<\/strong><\/em><\/h4>\n\n\n<hr>\n<p style=\"text-align: justify;\"><strong>Q-01)<\/strong> <em><strong>(Fuvest SP\/2022) <\/strong><\/em>Duas esferas de massa <em>\ud835\udc5a<\/em>, ambas carregadas eletricamente com a mesma carga <em>\ud835\udc5e<\/em>, est\u00e3o localizadas nas extremidades de fios isolantes, de comprimento <em>\ud835\udc3f<\/em>, presos ao teto, e formam o arranjo est\u00e1tico mostrado na figura.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"686\" height=\"391\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/eletro001.png\" alt=\"\" class=\"wp-image-901\" style=\"width:437px;height:249px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/eletro001.png 686w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/eletro001-600x342.png 600w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/eletro001-300x171.png 300w\" sizes=\"auto, (max-width: 686px) 100vw, 686px\" \/><\/figure>\n<\/div>\n\n<p style=\"text-align: justify;\">a) Fa\u00e7a um diagrama de corpo livre da esfera 1, indicando <u>todas<\/u> as for\u00e7as que atuam sobre ela.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E1.png\" alt=\"\" width=\"167\" height=\"158\" class=\"wp-image-902 alignnone\"><\/p>\n<p style=\"text-align: justify;\">b) Determine a raz\u00e3o &nbsp;em termos do comprimento <em>\ud835\udc3f<\/em> dos fios, da acelera\u00e7\u00e3o da gravidade <em>\ud835\udc54<\/em> e da constante eletrost\u00e1tica do v\u00e1cuo <em>\ud835\udc58<\/em>.<\/p>\n<p style=\"text-align: justify;\">c) Considere que as mesmas esferas s\u00e3o desconectadas dos fios e conectadas \u00e0s extremidades de uma mola de constante el\u00e1stica igual a 50 N\/m. O conjunto \u00e9 deixado sobre uma superf\u00edcie isolante e sem atrito, atingindo o equil\u00edbrio quando a for\u00e7a el\u00e9trica entre elas \u00e9 de 0,1 N. Nessas condi\u00e7\u00f5es, qual ser\u00e1 o valor da energia armazenada na mola?<\/p>\n<p style=\"text-align: justify;\"><strong>Note e adote<\/strong>:<\/p>\n<p style=\"text-align: justify;\">Despreze as dimens\u00f5es das esferas frente ao comprimento dos fios.<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-f74ade90\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\">a) Observe a figura abaixo onde s\u00e3o representadas todas as for\u00e7as atuantes.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"283\" height=\"327\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E2.png\" alt=\"\" class=\"wp-image-905\" style=\"width:176px;height:203px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E2.png 283w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E2-260x300.png 260w\" sizes=\"auto, (max-width: 283px) 100vw, 283px\" \/><figcaption class=\"wp-element-caption\">Tra\u00e7\u00e3o, For\u00e7a eletrost\u00e1tica e Peso.<\/figcaption><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">b) Voltando ao diagrama de corpo livre do item a), observamos que o peso \u00e9 igual \u00e0 for\u00e7a eletrost\u00e1tica. Sendo assim:<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"547\" height=\"298\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E3.png\" alt=\"\" class=\"wp-image-906\" style=\"width:308px;height:168px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E3.png 547w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E3-300x163.png 300w\" sizes=\"auto, (max-width: 547px) 100vw, 547px\" \/><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">Da trigonometria e da situa\u00e7\u00e3o de equil\u00edbrio de for\u00e7as, temos:<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"398\" height=\"322\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4.png\" alt=\"\" class=\"wp-image-907\" style=\"width:237px;height:192px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4.png 398w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4-300x243.png 300w\" sizes=\"auto, (max-width: 398px) 100vw, 398px\" \/><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) Note que foram dadas <em>F<sub>el\u00e9trica<\/sub> = 0,1 N<\/em>&nbsp;e <em>K = 50 N\/m<\/em>&nbsp;para, respectivamente, a for\u00e7a el\u00e9trica e para a constante el\u00e1stica da mola. Como o sistema est\u00e1 em equil\u00edbrio sobre uma superf\u00edcie horizontal, isolante e sem atrito, ent\u00e3o a for\u00e7a el\u00e9trica \u00e9 igual \u00e0 el\u00e1stica (<em>F<sub>el\u00e1stica<\/sub> = 0,1 N<\/em>). A energia armazenada na mola \u00e9 dada pela equa\u00e7\u00e3o <em>E<sub>mola<\/sub> = \u00bd.(k.x<sup>2<\/sup>)<\/em>. Da for\u00e7a el\u00e1stica extra\u00edmos o valor de <em>x = 2.10 <sup>-3<\/sup><\/em> metros e dessa forma podemos calcular a energia <em>E<sub>mola<\/sub> = 1.10 <sup>-4<\/sup> J<\/em>oules.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p style=\"text-align: justify;\"><strong>Q-02)<\/strong><span>&nbsp;<\/span><em><strong>(Fuvest SP\/2021) <\/strong><\/em>Dois bal\u00f5es negativamente carregados s\u00e3o utilizados para induzir cargas em latas met\u00e1licas, alinhadas e em contato, que, inicialmente, estavam eletricamente neutras.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"592\" height=\"252\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4-1.png\" alt=\"\" class=\"wp-image-917\" style=\"width:385px;height:164px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4-1.png 592w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E4-1-300x128.png 300w\" sizes=\"auto, (max-width: 592px) 100vw, 592px\" \/><\/figure>\n<\/div>\n\n<p>Conforme mostrado na figura, os bal\u00f5es est\u00e3o pr\u00f3ximos, mas jamais chegam a tocar as latas. Nessa configura\u00e7\u00e3o, as latas 1, 2 e 3 ter\u00e3o, respectivamente, carga total:<\/p>\n<p><strong>Note e adote<\/strong>:<\/p>\n<p>O contato entre dois objetos met\u00e1licos permite a passagem de cargas el\u00e9tricas entre um e outro.<\/p>\n<p style=\"text-align: justify;\">Suponha que o ar no entorno seja um isolante perfeito.<\/p>\n<p>a) 1: zero; 2: negativa; 3: zero.<\/p>\n<p>b) 1: positiva; 2: zero; 3: positiva.<\/p>\n<p>c) 1: zero; 2: positiva; 3: zero.<\/p>\n<p>d) 1: positiva; 2: negativa; 3: positiva.<\/p>\n<p>e) 1: zero; 2: zero; 3: zero.<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-88b39e40\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (D)<\/strong>&nbsp;\u2013 Observar que as latas met\u00e1licas por serem condutoras, se comportam como um \u00fanico corpo condutor, permitindo o fluxo de cargas ao longo delas.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p style=\"text-align: justify;\"><strong>Q-03)<\/strong><span>\u00a0<\/span><em><strong>(Famema SP\/2021) <\/strong><\/em>Em determinado meio, uma carga el\u00e9trica q \u00e9 colocada a uma dist\u00e2ncia de <em>1,2.10<sup>\u20132<\/sup> m<\/em> de outra carga Q, ambas pontuais. A essa dist\u00e2ncia, a carga q \u00e9 submetida a uma for\u00e7a repulsiva de intensidade <em>20 N<\/em>. Se a carga q for reposicionada a <em>0,4.10<sup>\u20132<\/sup> m<\/em> da carga Q no mesmo meio, a for\u00e7a repulsiva entre as cargas ter\u00e1 intensidade de<\/p>\n<p>a) 360 N.<\/p>\n<p>b) 480 N.<\/p>\n<p>c) 180 N.<\/p>\n<p>d) 520 N.<\/p>\n<p>e) 660 N.<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-d663a4e7\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (C)<\/strong>&nbsp;\u2013 Podemos comparar as duas situa\u00e7\u00f5es dividindo as for\u00e7as final e inicial, e aplicando a lei de Coulomb.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p style=\"text-align: justify;\"><strong>Q-04)<\/strong><span>\u00a0<\/span><em><strong>(Fuvest SP\/2019) <\/strong><\/em>Tr\u00eas pequenas esferas carregadas com carga positiva <em>Q<\/em> ocupam os v\u00e9rtices de um tri\u00e2ngulo, como mostra a figura. Na parte interna do tri\u00e2ngulo, est\u00e1 afixada outra pequena esfera, com carga negativa <em>q<\/em>. As dist\u00e2ncias dessa carga \u00e0s outras tr\u00eas podem ser obtidas a partir da figura.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"730\" height=\"585\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/F01.png\" alt=\"\" class=\"wp-image-931\" style=\"width:341px;height:273px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/F01.png 730w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/F01-600x481.png 600w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/F01-300x240.png 300w\" sizes=\"auto, (max-width: 730px) 100vw, 730px\" \/><\/figure>\n<\/div>\n\n<p>Sendo <em>Q <\/em>= 2.10<sup>\u2013 4<\/sup><em>C, q=<\/em> \u20132.10<sup>\u20135<\/sup>C e <em>d<\/em> = 6m, a for\u00e7a el\u00e9trica resultante sobre a carga <em>q<\/em><\/p>\n<p>a) \u00e9 nula.<\/p>\n<p>b) tem dire\u00e7\u00e3o do eixo y, sentido para baixo e m\u00f3dulo 1,8 N.<\/p>\n<p>c) tem dire\u00e7\u00e3o do eixo y, sentido para cima e m\u00f3dulo 1,0 N.<\/p>\n<p>d) tem dire\u00e7\u00e3o do eixo y, sentido para baixo e m\u00f3dulo 1,0 N.<\/p>\n<p>e) tem dire\u00e7\u00e3o do eixo y, sentido para cima e m\u00f3dulo 0,3 N.<\/p>\n<p><strong>Note e adote:<\/strong><\/p>\n<p><strong>A constante k<sub>0<\/sub> da lei de Coulomb vale 9.10<sup>9<\/sup> N.m<sup>2<\/sup>\/C<sup>2<\/sup><\/strong><\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-1b0aafae\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (E)<\/strong>.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p><strong>Q-05)<\/strong><span>\u00a0<\/span><em><strong>(IME RJ\/2019)\u00a0<\/strong><\/em><\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"687\" height=\"536\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/I01.png\" alt=\"\" class=\"wp-image-934\" style=\"width:541px;height:422px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/I01.png 687w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/I01-600x468.png 600w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/I01-300x234.png 300w\" sizes=\"auto, (max-width: 687px) 100vw, 687px\" \/><\/figure>\n<\/div>\n\n<p style=\"text-align: justify;\">A figura mostra uma estrutura composta pelas barras AB, AC, AD e CD e BD articuladas em suas extremidades. O apoio no ponto A impede os deslocamentos nas dire\u00e7\u00f5es x e y, enquanto o apoio no ponto C impede o deslocamento apenas na dire\u00e7\u00e3o x. No ponto D dessa estrutura encontra-se uma part\u00edcula el\u00e9trica de carga positiva q. Uma part\u00edcula el\u00e9trica de carga positiva Q encontra-se posicionada no ponto indicado na figura. Uma for\u00e7a de 10 N \u00e9 aplicada no ponto B, conforme indicada na figura. Para que a for\u00e7a de rea\u00e7\u00e3o no ponto C seja zero, o produto q.Q deve ser igual a<\/p>\n<p><strong>Observa\u00e7\u00e3o<\/strong>:<\/p>\n<ul>\n<li>as barras e part\u00edculas possuem massa desprez\u00edvel; e<\/li>\n<li>as dist\u00e2ncias nos desenhos est\u00e3o representadas em metros.<\/li>\n<\/ul>\n<p><strong>Dado<\/strong>:<\/p>\n<ul>\n<li>constante eletrost\u00e1tica do meio: k.<\/li>\n<\/ul>\n<p>a) 1250\/7k<\/p>\n<p>b) 125\/70k<\/p>\n<p>c) 7\/1250k<\/p>\n<p>d) 1250\/k<\/p>\n<p>e) k\/1250<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-8b7db946\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (A)<\/strong>.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p style=\"text-align: justify;\"><strong>Q-06)<\/strong><em><strong>(UERJ\/2018) <\/strong><\/em>O esquema abaixo representa as esferas met\u00e1licas A e B, ambas com massas de 10<sup>\u20133<\/sup> kg e carga el\u00e9trica de m\u00f3dulo igual a 10<sup>\u20136<\/sup> C. As esferas est\u00e3o presas por fios isolantes a suportes, e a dist\u00e2ncia entre elas \u00e9 de 1 m.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"888\" height=\"231\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E02.png\" alt=\"\" class=\"wp-image-936\" style=\"width:625px;height:162px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E02.png 888w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E02-600x156.png 600w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E02-300x78.png 300w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/E02-768x200.png 768w\" sizes=\"auto, (max-width: 888px) 100vw, 888px\" \/><\/figure>\n<\/div>\n\n<p style=\"text-align: justify;\">Admita que o fio que prende a esfera A foi cortado e que a for\u00e7a resultante sobre essa esfera corresponde apenas \u00e0 for\u00e7a de intera\u00e7\u00e3o el\u00e9trica. Calcule a acelera\u00e7\u00e3o, em m\/s<sup>2<\/sup>, adquirida pela esfera A imediatamente ap\u00f3s o corte do fio.<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-00eac3f2\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\">Pode-se aplicar a 2\u00aa Lei de Newton para o c\u00e1lculo da acelera\u00e7\u00e3o, conforme o desenvolvimento abaixo.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"447\" height=\"288\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/R01.png\" alt=\"\" class=\"wp-image-939\" style=\"width:270px;height:174px\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/R01.png 447w, https:\/\/proexatas.com\/wp-content\/uploads\/2023\/05\/R01-300x193.png 300w\" sizes=\"auto, (max-width: 447px) 100vw, 447px\" \/><\/figure>\n<\/div><\/div><\/details>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n<p style=\"text-align: justify;\"><strong>Q-07)<\/strong><em><strong>(PUC-SP) <\/strong><\/em>Uma placa retangular de madeira Pinus elliottii, cuja densidade \u00e9 igual a 0,5g\/cm<sup>3<\/sup>, possui as seguintes dimens\u00f5es de arestas: 20cm x 40cm x 8cm. Ela encontra-se boiando em equil\u00edbrio no interior de uma cuba preenchida com benzeno, cuja densidade \u00e9 de 0,9 g\/cm<sup>3<\/sup>. Depois de um certo instante, no centro da superf\u00edcie emersa da placa de madeira, \u00e9 fixada uma pequenina esfera met\u00e1lica, de massa desprez\u00edvel e eletrizada com carga q<sub>1<\/sub> = \u20131,0 microcoulombs. Ent\u00e3o, o sistema \u201cmadeira+esfera\u201d \u00e9 posicionado abaixo de um outro sistema formado por uma pequenina esfera met\u00e1lica, id\u00eantica \u00e0quela fixada na madeira, um fio isolante e um suporte tamb\u00e9m isolante. Essa segunda esferinha met\u00e1lica est\u00e1 eletrizada com carga q<sub>2<\/sub> = +20,0 microcoulombs. A dist\u00e2ncia entre os centros das esferas, consideradas pontuais, \u00e9 de 10cm, conforme indica a figura.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"885\" height=\"446\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image.png\" alt=\"\" class=\"wp-image-1030\" style=\"width:634px;height:auto\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image.png 885w, https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-600x302.png 600w, https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-300x151.png 300w, https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-768x387.png 768w\" sizes=\"auto, (max-width: 885px) 100vw, 885px\" \/><\/figure>\n<\/div>\n\n<p style=\"text-align: justify;\">Ap\u00f3s alguns segundos, verifica-se o equil\u00edbrio dos sistemas. Nas condi\u00e7\u00f5es de equil\u00edbrio, determine a raz\u00e3o aproximada, em porcentagem (%), entre os volumes imersos da placa de madeira com e sem a presen\u00e7a das esferinhas met\u00e1licas:<\/p>\n<p>a) 55.<\/p>\n<p>b) 50.<\/p>\n<p>c) 45.<\/p>\n<p>d) 35.<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-93651f0b\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (C)<\/strong>.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n<p style=\"text-align: justify;\">Tr\u00eas part\u00edculas eletrizadas com cargas positivas, q<sub>A<\/sub> = q<sub>B<\/sub> = q<sub>C<\/sub>, est\u00e3o fixas em tr\u00eas v\u00e9rtices de um cubo, conforme a figura.<\/p>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"541\" height=\"321\" src=\"https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-4.png\" alt=\"\" class=\"wp-image-1037\" style=\"width:429px;height:auto\" srcset=\"https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-4.png 541w, https:\/\/proexatas.com\/wp-content\/uploads\/2025\/02\/image-4-300x178.png 300w\" sizes=\"auto, (max-width: 541px) 100vw, 541px\" \/><\/figure>\n<\/div>\n\n<p style=\"text-align: justify;\">Sendo F<sub>AB<\/sub> a intensidade da for\u00e7a de repuls\u00e3o entre q<sub>A<\/sub> e q<sub>B<\/sub>, F<sub>BC<\/sub> a intensidade da for\u00e7a de repuls\u00e3o entre q<sub>B<\/sub> e q<sub>C<\/sub>, e F<sub>AC<\/sub> a intensidade da for\u00e7a de repuls\u00e3o entre q<sub>A<\/sub> e q<sub>C<\/sub>, \u00e9 correto afirmar que<\/p>\n<p>a) F<sub>AB<\/sub> = \u221a2.F<sub>BC<\/sub> = \u221a3.F<sub>AC<\/sub><\/p>\n<p>b)\u00a0<\/p>\n\n\n<div id=\"wp-block-themeisle-blocks-accordion-a93a92b3\" class=\"wp-block-themeisle-blocks-accordion exclusive\">\n<details class=\"wp-block-themeisle-blocks-accordion-item\"><summary class=\"wp-block-themeisle-blocks-accordion-item__title\"><div>Visualizar alternativa correta \/ coment\u00e1rio da quest\u00e3o<\/div><\/summary><div class=\"wp-block-themeisle-blocks-accordion-item__content\">\n<p class=\"wp-block-paragraph\"><strong>Alternativa correta (E)<\/strong>.<\/p>\n<\/div><\/details>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Q-01) (Fuvest SP\/2022) Duas esferas de massa \ud835\udc5a, ambas carregadas eletricamente com a mesma carga \ud835\udc5e, est\u00e3o localizadas nas extremidades de fios isolantes, de comprimento \ud835\udc3f, presos ao teto, e formam o arranjo est\u00e1tico mostrado na figura. a) Fa\u00e7a um diagrama de corpo livre da esfera 1, indicando todas as [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_themeisle_gutenberg_block_has_review":false,"footnotes":""},"class_list":["post-895","page","type-page","status-publish","hentry"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/pages\/895","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/comments?post=895"}],"version-history":[{"count":22,"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/pages\/895\/revisions"}],"predecessor-version":[{"id":1248,"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/pages\/895\/revisions\/1248"}],"wp:attachment":[{"href":"https:\/\/proexatas.com\/index.php\/wp-json\/wp\/v2\/media?parent=895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}