{"id":58162,"date":"2024-04-16T01:14:39","date_gmt":"2024-04-16T01:14:39","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=58162"},"modified":"2024-04-16T01:14:39","modified_gmt":"2024-04-16T01:14:39","slug":"two-plates-of-a-parallel-plate-capacitor-after-being-charged-from-a-constant-voltage-source-are-separated-apart-by-means-of-insulated-handles-then-the","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-plates-of-a-parallel-plate-capacitor-after-being-charged-from-a-constant-voltage-source-are-separated-apart-by-means-of-insulated-handles-then-the\/","title":{"rendered":"Two plates of a parallel plate capacitor after being charged from a constant voltage source are separated apart by means of insulated handles, then the"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a970db5447d6\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    Voltage across the plates increases                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    Voltage across the plates decreases                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"true\">\r\n                    Charge on the capacitor decreases                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    Charge on the capacitor increases                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The correct <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> answer is: <strong>C. Charge on the capacitor decreases<\/strong>.<\/p>\n<p>When the plates of a parallel plate capacitor are separated, the electric field between the plates decreases. This is because the charge on the plates is spread out over a larger area. The electric field is proportional to the charge, so the decrease in electric field means that the charge on the plates must also decrease.<\/p>\n<p>Here is a more detailed explanation of each option:<\/p>\n<ul>\n<li>Option A: Voltage across the plates increases. This is incorrect because the voltage across the plates is determined by the charge <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div> on the plates and the capacitance of the capacitor. The capacitance of the capacitor is a constant, so the voltage across the plates cannot change when the charge on the plates changes.<\/li>\n<li>Option B: Voltage across the plates decreases. This is incorrect for the same reason as Option A.<\/li>\n<li>Option C: Charge on the capacitor decreases. This is correct because the electric field between the plates decreases when the plates are separated. The electric field is proportional to the charge, so the decrease in electric field means that the charge on the plates must also decrease.<\/li>\n<li>Option D: Charge on the capacitor increases. This is incorrect because the charge on the plates decreases when the plates are separated.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Subscribe on YouTube<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[971],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Two plates of a parallel plate capacitor after being charged from a constant voltage source are separated apart by means of insulated handles, then the<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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