Question

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​

Answer 1

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

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

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

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

Answer 2

Answer:

$\large\boxed{\sf{1.01\;N}}$

Explanation:

It's being given that there is a parallel plate capacitor.

Charge on capacitor = 6 μC = $6\times {10}^{-6}\:C$

Area of the plates = $2\:{m}^{2}$

To find the force between the plates:

We know that, for a parallel plate capacitor,

$\large \boxed{ \red{E = \dfrac{q}{2A \epsilon_{o}} }}$

where, the respective meanings of,

• E = electric field
• q = charge on capacitor
• A = area between the plates
• $\epsilon_{o}=8.85\times{10}^{-12}\;N{m}^{2}{C}^{-2}$

Also, we know that,

$\large\boxed{\red{F=qE}}$

where F is the force required.

Therefore, we have,

$= > F = \frac{ {q}^{2} }{2A \epsilon_{o} } \\ \\ = > F = \frac{ {(6 \times {10}^{ - 6} )}^{2} }{2 \times 2 \times 8.85 \times {10}^{ - 12} } \\ \\ = > F= \frac{36 \times {10}^{ - 12} }{4 \times 8.85 \times {10}^{ - 12} } \\ \\ = > F = \frac{36}{4 \times 8.85} \\ \\ = > F = \frac{9}{8.85} \\ \\ = > F = 1.01 \: N$

Hence, force of attraction is 1.01 N