","comment":{"@type":"Comment","text":" Capacitance of an isolated sphere is $C=4\\pi {{\\in }_{0}}R$. Thus the capacitance depends directly on the radius of the sphere. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{C}_{1}} > {{C}_{2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $${{C}_{1}} = {{C}_{2}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$Zero$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{C}_{2}} > {{C}_{1}}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$C=4\\pi {{\\in }_{0}}R \\\\ \\Rightarrow C\\propto R \\\\ Given\\text{ }{{R}_{2}}>{{R}_{1}} \\\\ \\therefore {{C}_{2}}>{{C}_{1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Capacitors Quiz 1","text":"A capacitor of $$C=3\\mu F$$ is charged with a battery of $$5$$ volt. The plate separation is $$d$$.When it achieves maximum charge, the battery is disconnected and the plates of the capacitor are pulled apart to distance $$''2d''$$. Now the charge and potential on each capacitor: $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Charge stored in capacitor is $Q=CV$ . On disconnecting the battery, charge stored in the capacitor remains constant to obey the law of conservation of charge for an isolated system. Also for a parallel plate capacitor, $C\\propto \\dfrac{1}{d}$, where $d$ is separation between plates of capacitor. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" charge gets halved, potential remains constant","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" charge remains constant, potential remains constant too","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" charge gets doubled, potential remains constant","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" charge remains constant, potential gets doubled","position":2,"answerExplanation":{"@type":"Comment","text":" As the system is isolated, the charge remains the same to obey law of conservation of charge. $$ \\\\ $$ For a parallel plate capacitor, $$C=\\dfrac{{{\\varepsilon }_{0}}A}{d}$$. $$ \\\\ $$ Thus, $$ \\text{ C}\\propto \\dfrac{1}{d} $$ $$ \\\\ \\dfrac{{{\\text{C}}_{1}}}{{{\\text{C}}_{2}}}=\\dfrac{{{d}_{2}}}{{{d}_{1}}} \\\\ \\Rightarrow \\dfrac{\\text{C}}{\\text{C }\\!\\!'\\!\\!\\text{ }}=\\dfrac{2d}{d} \\\\ \\Rightarrow C'=\\dfrac{C}{2} \\\\ $$ So as width ādā gets doubled, capacitance will become half. $$ \\\\ $$ For a capacitor, $$C=\\dfrac{Q}{V}$$ $$ \\\\ \\text{C}\\propto \\dfrac{1}{V} \\\\ $$ Thus, the potential increases by two times if C is halved.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Capacitors Quiz 1","text":" Two identical capacitors of capacitance $$''C''$$ are connected in series to a battery of emf $$''E''$$. The first capacitor is filled completely with a dielectric of dielectric constant $$3$$, while in the second capacitor a metal slab of thickness $$\\dfrac{1}{3}$$ times the separation of plates the capacitor is inserted between the plates. The amount of charge stored after the insertion of dielectric and metal slab is $$ \\\\ $$ ","comment":{"@type":"Comment","text":" For a capacitor of plates area $A$ and plate separation $d$, the capacitance after insertion of a metal slab of $t$ thickness is $C'=\\dfrac{{{\\varepsilon }_{0}}A}{d.t}$ and if filled with dielectric capacitance is $C'=KC$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" thrice that of charge stored initially","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" half of charge stored initially","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" four times of that of charge stored initially","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" twice that of charge stored initially","position":0,"answerExplanation":{"@type":"Comment","text":" If the initial separation of plates be $$d$$ and the area of a plate be $$A$$, then $$ \\\\ C=\\dfrac{{{\\varepsilon }_{0}}A}{d} $$ $$ \\\\ $$ After inserting a dielectric in the first capacitor, the new capacitance is $${{C}_{1}}=KC=3C$$ $$ \\\\ $$ Similarly, after inserting a metal slab of $$\\dfrac{d}{3}$$ thickness, the capacitance of the second capacitor is $$ \\\\ {{C}_{2}}=\\dfrac{{{\\varepsilon }_{0}}A}{d-\\dfrac{d}{3}}=\\dfrac{3}{2}\\dfrac{{{\\varepsilon }_{0}}A}{d}=\\dfrac{3}{2}C \\\\ $$ As $${{C}_{1}}$$ and $${{C}_{2}}$$ are in series, their equivalent is $$ \\\\ C{{'}_{\\text{series}}}=\\dfrac{{{C}_{1}}\\times {{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\\dfrac{3C\\times \\dfrac{3}{2}C}{3C+\\dfrac{3}{2}C}=C \\\\ $$ Initially, without the dielectric and metal slab, the equivalent capacitance is $$ \\\\ {{C}_{\\text{series}}}=\\dfrac{C\\times C}{C+C}=\\dfrac{C}{2} \\\\ $$ For a capacitor, $$Q=CV$$ $$ \\\\ \\text{If }V\\text{ is constant,} \\\\ Q\\propto C \\\\ \\dfrac{{{Q}_{\\text{final}}}}{{{Q}_{\\text{initial}}}}=\\dfrac{C{{'}_{\\text{series}}}}{{{C}_{\\text{series}}}}=\\dfrac{C}{\\dfrac{C}{2}}=2 \\\\ \\therefore {{Q}_{\\text{final}}}=2{{Q}_{\\text{initial}}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Capacitors Quiz 1","text":" In a parallel plate capacitor of area $$''A''$$ and separation $$''d''$$, two dielectric slabs of width $$\\dfrac{d}{3}$$ each and dielectric constants $${{k}_{1}}$$ and $${{k}_{2}}$$ respectively, are inserted. The first dielectric is inserted in contact with the first plate, and the second dielectric is placed in contact with the other plate. Calculate the new capacitance of the capacitor. $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Each part of the capacitor with a different dielectric medium can be considered as an individual capacitor. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3{{\\varepsilon }_{0}}A}{d}\\left( {{k}_{1}}+{{k}_{2}}+1 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3{{\\varepsilon }_{0}}A}{d}\\left( \\dfrac{{{k}_{1}}\\times {{k}_{2}}}{{{k}_{1}}+{{k}_{2}}+1} \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3{{\\varepsilon }_{0}}A}{d}\\left( 1+\\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3{{\\varepsilon }_{0}}A}{d}\\left( \\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{2}}+{{k}_{1}}{{k}_{2}}+{{k}_{2}}} \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":" Each part of the capacitor filled with a different dielectric can be considered as an individual capacitor $$ \\\\ \\text{Capacitance }{{C}_{1}}={{k}_{1}}\\dfrac{{{\\varepsilon }_{0}}A}{\\dfrac{d}{3}}=3{{k}_{1}}\\dfrac{{{\\varepsilon }_{0}}A}{d} \\\\ \\text{Capacitance }{{C}_{2}}=\\dfrac{{{\\varepsilon }_{0}}A}{\\dfrac{d}{3}}=3\\dfrac{{{\\varepsilon }_{0}}A}{d} \\\\ \\text{Capacitance }{{C}_{3}}={{k}_{2}}\\dfrac{{{\\varepsilon }_{0}}A}{\\dfrac{d}{3}}=3{{k}_{2}}\\dfrac{{{\\varepsilon }_{0}}A}{d} \\\\ $$ $${{C}_{1}},{{C}_{2}}$$ and $${{C}_{3}}$$ are in continuous sequence between $$A$$ and $$B$$. Thus they are in series. For a series combination, the equal capacitance is $$ \\\\ \\dfrac{1}{{{C}_{series}}}=\\dfrac{1}{{{C}_{1}}}+\\dfrac{1}{{{C}_{2}}}+\\dfrac{1}{{{C}_{3}}}=\\dfrac{d}{3{{k}_{1}}{{\\varepsilon }_{0}}A}+\\dfrac{d}{3{{\\varepsilon }_{0}}A}+\\dfrac{d}{3{{k}_{2}}{{\\varepsilon }_{0}}A} \\\\ \\Rightarrow \\dfrac{1}{{{C}_{series}}}=\\dfrac{d}{3{{\\varepsilon }_{0}}A}\\left( \\dfrac{1}{{{k}_{1}}}+1+\\dfrac{1}{k2} \\right) \\\\ \\therefore {{C}_{series}}=\\dfrac{3{{\\varepsilon }_{0}}A}{d}\\left( \\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{2}}+{{k}_{1}}{{k}_{2}}+{{k}_{1}}} \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Capacitors Quiz 1","text":"If in a given circuit, find the ratio of equivalent capacitance when switch is open and closed respectively.$${{C}_{1}}=2\\mu F$$, $${{C}_{2}}=4\\mu F$$, $${{C}_{3}}=3\\mu F$$, $${{C}_{4}}=5\\mu F$$ $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Use ${{C}_{eq}}=\\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ for series, ${{C}_{eq}}={{C}_{1}}+{{C}_{2}}$ for parallel combination, where ${{C}_{equivalent}}$ is equivalent capacitance. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$1:1$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{12\\times 45}{77\\times 7}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{25}{12}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{77\\times 7}{45\\times 12}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ $$ $$ \\\\ $$ $$ \\\\ $$ $$ \\\\ $$ $$ \\\\ \\text{When switch is open, the circuit looks like as shown above} \\\\ {{\\text{C}}_{1}}\\text{ and }{{\\text{C}}_{2}}\\text{ are in series } \\\\ \\text{Also }{{\\text{C}}_{3}}\\text{ and }{{\\text{C}}_{4}}\\text{ are in series}\\text{.} \\\\ \\text{For series combination, the equivalent capacitance is } \\\\ {{\\text{C}}_{series}}=\\dfrac{{{C}_{1}}\\times {{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\\text{=}\\dfrac{\\text{2}\\times \\text{4}}{2+4}\\text{=}\\dfrac{8}{6}\\text{=}\\dfrac{\\text{4}}{3}\\text{ }\\mu \\text{F, for }{{\\text{C}}_{1}}\\text{ and }{{\\text{C}}_{2}} \\\\ \\text{Similarly for }{{\\text{C}}_{3}}\\text{ and }{{\\text{C}}_{4}}\\text{, the equivalent is =}\\dfrac{3\\times 5}{3+5}\\text{=}\\dfrac{15}{8}\\text{ }\\mu \\text{F} \\\\ These\\text{ two equivalent capacitors are in parallel with each other}\\text{.} \\\\ \\text{For parallel combination, the equivalent is } \\\\ {{C}_{parallel}}=C+C'=\\dfrac{\\text{4}}{3}+\\dfrac{15}{8}=\\dfrac{77}{24}\\text{ }\\mu \\text{F} \\\\\\text{For figure 2, after switch closes, } \\\\ {{\\text{C}}_{1}}\\text{ and }{{\\text{C}}_{3}}\\text{ are in parallel with each other as well as }{{\\text{C}}_{2}}\\text{ and }{{\\text{C}}_{4}}. \\\\ \\text{Equivalent of }{{\\text{C}}_{1}}\\text{ and }{{\\text{C}}_{3}}\\text{ =2+3=5 }\\mu \\text{F} \\\\ \\text{Equivalent of }{{\\text{C}}_{2}}\\text{ and }{{\\text{C}}_{4}}\\text{ =4+5=9 }\\mu \\text{F} \\\\ \\text{These two capacitors are in parallel} \\\\ \\text{. So their equivalent is} \\\\ {{\\text{C}}_{series}}=\\dfrac{5\\times 9}{5+9}=\\dfrac{45}{14}\\text{ }\\mu \\text{F } \\\\ \\text{The ratio is =}\\dfrac{77}{24}\\times \\dfrac{14}{45} \\\\ \\text{The ratio is =}\\dfrac{77\\times 7}{45\\times 12}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}