{"id":4693,"date":"2021-09-29T15:23:12","date_gmt":"2021-09-29T09:53:12","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=4693"},"modified":"2024-09-30T08:10:40","modified_gmt":"2024-09-30T02:40:40","slug":"five-conducting-parallel-plates","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/five-conducting-parallel-plates\/","title":{"rendered":"Five Conducting Parallel Plates"},"content":{"rendered":"\t\t
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Consider a system of five conducting parallel plates (large plates) of area $A$. The charge on plates $1$, $2$, $3$, $4$ and $5$ is $Q$,\u00a0$2Q$,\u00a0<\/span>$3Q$,\u00a0<\/span>$4Q$ and\u00a0<\/span>$5Q$ respectively. Initially the two switches are opened. Assume that that distance between the plates, $d$, is very small compared to the dimensions of the plate.<\/span><\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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So, before switches $S_1$ and $S_2$ are closed, let’s determine
$(1)$ The charge on surface $b$ and
$(2)$ The potential difference between plate $2$ and $3$, i.e. $V_2 – V_3$<\/p>

So let’s begin with finding the charge on surface $b$ and assume that the charge densities on respective surfaces are as shown\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Two governing principles are:\u00a0
(i) Electric field inside all the metallic plates should be zero
(ii) Net charge on the plates (excess + induced) should be equal to the excess charge as stated in the problem statement<\/p>

So, electric field inside plate $1$ should be zero implies
$\\sigma_a = \\sigma_b + \\sigma_c + \\sigma_d + \\sigma_e + \\sigma_f +\\sigma_g + \\sigma_h + \\sigma_i + \\sigma_j$ $\\cdots$ (1)<\/p>

electric field inside plate $2$ should be zero gives us
$\\sigma_a + \\sigma_b + \\sigma_c = \\sigma_d + \\sigma_e + \\sigma_f +\\sigma_g + \\sigma_h + \\sigma_i + \\sigma_j$ $\\cdots$ (2)<\/p>

electric field inside plate $3$ should be zero gives us
$\\sigma_a + \\sigma_b + \\sigma_c + \\sigma_d + \\sigma_e = \\sigma_f +\\sigma_g + \\sigma_h + \\sigma_i + \\sigma_j$ $\\cdots$ (3)<\/p>

electric field inside plate $4$ should be zero gives us
$\\sigma_a + \\sigma_b + \\sigma_c + \\sigma_d + \\sigma_e + \\sigma_f +\\sigma_g = \\sigma_h + \\sigma_i + \\sigma_j$ $\\cdots$ (4)<\/p>

electric field inside plate $5$ should be zero gives us
$\\sigma_a + \\sigma_b + \\sigma_c + \\sigma_d + \\sigma_e + \\sigma_f +\\sigma_g + \\sigma_h + \\sigma_i = \\sigma_j$ $\\cdots$ (5)<\/p>

And conservation of charge would give us
$\\sigma_a + \\sigma_b = \\cfrac{Q}{A}$ $\\cdots$ (6)
$\\sigma_c + \\sigma_d = \\cfrac{2Q}{A}$ $\\cdots$ (7)
$\\sigma_e + \\sigma_f = \\cfrac{3Q}{A}$ $\\cdots$ (8)
$\\sigma_g + \\sigma_h = \\cfrac{4Q}{A}$ $\\cdots$ (9)
$\\sigma_i + \\sigma_j = \\cfrac{5Q}{A}$ $\\cdots$ (10)<\/p>

Adding (1) and (5), we get $\\sigma_a = \\sigma_j$<\/p>

Subtracting (2) from (1) we get $\\sigma_b = – \\sigma_c$<\/p>

Subtracting (3) from (2) we get $\\sigma_d = – \\sigma_e$<\/p>

Subtracting (4) from (3) we get $\\sigma_f = – \\sigma_g$<\/p>

Subtracting (5) from (4) we get $\\sigma_h = – \\sigma_i$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Now to find surface charge density (or total charge) on surface ‘$b$’, equations (1) and (7) to (10) give us\u00a0<\/p>

$\\sigma_a – \\sigma_b = \\cfrac{14Q}{A}$ $\\cdots$ (11)<\/p>

From (6) and (11) we get
$\\sigma_a = \\cfrac{7.5Q}{A}$
or $Q_a = 7.5Q$
$Q_b = -6.5Q$
$Q_c = 6.5Q$
$Q_d = -4.5Q$
$Q_e = 4.5Q$
$Q_f = -1.5Q$
$Q_g = 1.5Q$
$Q_h = 2.5Q$
$Q_i = -2.5Q$
$Q_j = 7.5Q$<\/p>

Note $Q_a = Q_b$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Potential difference between plates (2) and (3) would be:
$V_2 – V_3 = – \\int \\overrightarrow{E}.\\overrightarrow{dl}$ $= – \\cfrac{\\sigma_e}{\\epsilon_0}$ $= – \\cfrac{4.5Q}{\\epsilon_0 A}$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Now assume that both the switches $S_1$ and $S_2$ are closed simultaneously, and determine
$(3)$ The charge on surface $b$ and
$(4)$ The charge on surface $f$.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Equations (1) through (5) hold true in this scenario as well, which means, that\u00a0
$\\sigma_a = \\sigma_j$
$\\sigma_b = -\\sigma_c$
$\\sigma_d = -\\sigma_e$
$\\sigma_f = -\\sigma_g$
$\\sigma_h = -\\sigma_i$<\/p>

Now, total excess charge on plates 1 + 4 $=5Q$
& total excess charge on plates 3 + 5 $=8Q$<\/p>

So, $\\sigma_a + \\sigma_b + \\sigma_g + \\sigma_h = \\cfrac{5Q}{A}$
$\\sigma_c + \\sigma_d = \\cfrac{2Q}{A}$
$\\sigma_e + \\sigma_f + \\sigma_i + \\sigma_j = \\cfrac{8Q}{A}$<\/p>

Moreover, potential difference between plates 1 and 4 is zero and potential difference between 3 and 5 is zero, which means\u00a0
$\\sigma_b + \\sigma_d + \\sigma_f = 0$
& $\\sigma_f + \\sigma_h = 0$<\/p>

Solving we get $\\sigma_a = \\cfrac{7.5Q}{A}$
and $\\sigma_f = \\cfrac{3}{5} \\cfrac{Q}{A}$<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

Consider a system of five conducting parallel plates of large area A. The charge on plates is Q, 2Q, 3Q, 4Q and 5Q. Determine the charge on different surfaces.<\/p>\n","protected":false},"author":1,"featured_media":4689,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"0","ocean_second_sidebar":"0","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"0","ocean_custom_header_template":"0","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"0","ocean_menu_typo_font_family":"0","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"on","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"off","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"0","cybocfi_hide_featured_image":"yes","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"off","ocean_gallery_id":[],"footnotes":""},"categories":[16],"tags":[],"class_list":["post-4693","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jee-physics-chapter-wise-approach","entry","has-media"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4693","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/comments?post=4693"}],"version-history":[{"count":1,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4693\/revisions"}],"predecessor-version":[{"id":9437,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/4693\/revisions\/9437"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media\/4689"}],"wp:attachment":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media?parent=4693"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/categories?post=4693"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/tags?post=4693"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}