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two isolated plates of a capacitor has been given a charge of 6microcoulomb, the area of plates is 2 sq metre.Find approximate force of attraction between two plate​

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\large\sf\underline{ \red{\underline{Question }}}:- \: \\ \sf Two \: isolated \: plates \: of \: a \: capacitor \: has \: been \: given \: \\ \sf a \: <klux>CHARGE</klux> \: of \: 6 \mu \: c \: the \: area \: of \: plates \: is \: 2sq. \: <klux>METRE</klux> \\ \sf find \: approximate \: force \: of \: attraction \: between \\ \sf two \:p lates.

\large\sf\underline{ \orange{\underline{ Answer }}}:- \: \: \\ \\ \: \to \red{ \boxed{ \sf \green{F = 1.048 \: Newton \: }}} \: \\ \\ \large\sf\underline{ \green{\underline{ To Find \: }}}:- \: \: \\ \to \sf \: approximate \: force \: between \: plates

\large\sf\underline{ \red{\underline{ Explanation\: }}}:-

Given that :

  • Area between plates (A) = 2 sq. metre

  • Charge on plates \sf(Q) = 6 \mu C

\large\sf\underline{ \pink{\underline{ Solution \: }}}:- \:

We know that ,

\to \sf \red{E = } \green{\dfrac{Q}{2 A \epsilon_0 } \: } \\ \: \\ \because \sf \: F = EQ \\ \: \therefore \\ \\ \to \sf F = Q \times \dfrac{Q}{2 A \epsilon_0 } \\ \\ \to \sf F = \dfrac{ {Q}^{2} }{2 A \epsilon_0 } \: \\ \\ \red{\boxed{ \because \sf \green{ \epsilon_0 = 8.85} \times {10}^{ - 12} }} \\ \boxed{ \because \sf1 \mu \: C = {10}^{ - 6} \: C \: \: \: \: \: \: \: \: \: \: } \\ \bf puting \: the \: given \: values \\

\to \sf F \: = { \frac{ \orange{6 \times {10}^{ - 6} \times 6 \times {10}^{ - 6} } }{ \red{2 \times 2 \times 8.85 \times {10}^{ - 12} }}} \\ \\ \to \sf F = \frac{36 \times {10}^{ - 12} }{ \blue{4 \times 8.85 \times {10}^{ - 12} }} \\ \\ \to \sf \: F = \frac{36 \times 100}{4 \times 858} \\ \\ \to \sf F = \pink{ \frac{3600}{3432} }\\ \\ \to \red{ \boxed{ \sf \green{F = 1.048 \: Newton \: }}}

REQUIRED force between them is 1.048 N


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