Doubt in em

Hi brilliant,

I've got a problem in question number 3.143 of the book Problems in general physics by Irodov.

The question states,

A parallel-plate capacitor was lowered into water in a horizontal, with water filling up the gap between the plates d meter wide. Then a constant voltage \(V\) was applied to the capacitor. Find the water pressure increment in the gap. Take relative permittivity of water as \(\varepsilon \).

I solved it as follows:

  • Electrostatic pressure on a point in the capacitor before water was filled:

\({ P }_{ 1 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ \left( \frac { V }{ d } \right) }^{ 2 }\)

  • Electrostatic pressure on a point in the capacitor before water was filled:

\({ P }_{ 2 }=\frac { 1 }{ 2 } { \varepsilon \varepsilon }_{ 0 }{ E }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon \varepsilon }_{ 0 }{ \left( \frac { V }{ d } \right) }^{ 2 }\)

  • Therefore increase in pressure = \(\Delta P=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }\left( \varepsilon -1 \right) { \left( \frac { V }{ d } \right) }^{ 2 }\)

But the answer given in the solution is:

\(\Delta P=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }\varepsilon \left( \varepsilon -1 \right) { \left( \frac { V }{ d } \right) }^{ 2 }\)

Please check if my solution is right. Please correct me.

Note by Aditya Kumar
3 years, 1 month ago

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1 vote

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Yeah, dimensionally speaking, the given answer is wrong. Must be a typo.

Raj Magesh - 3 years, 1 month ago

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Can u verify my solution??

Aditya Kumar - 3 years, 1 month ago

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Wait a minute: does the question say that permittivity of water is \(\epsilon\) or that the relative permittivity of water is \(\epsilon\)?

Raj Magesh - 3 years, 1 month ago

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@Raj Magesh But your method is fine.

Raj Magesh - 3 years, 1 month ago

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@Raj Magesh Thanks for reviewing it!

Aditya Kumar - 3 years, 1 month ago

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@Raj Magesh Relative permittivity

Aditya Kumar - 3 years, 1 month ago

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This looks like a typo to me, unless I'm missing something.

Josh Silverman Staff - 3 years, 1 month ago

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Sir what can u say about my method? Is it right?

Aditya Kumar - 3 years, 1 month ago

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I find no problem with your approach. I think your work is correct.

Josh Silverman Staff - 3 years, 1 month ago

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@Josh Silverman I don't think this method is right - the charge distribution and hence the field will not remain same with and without the dielectric (water in this case). What do you think?

Rahul Sethi - 4 months, 1 week ago

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@Josh Silverman Thanks a lot sir!

Aditya Kumar - 3 years, 1 month ago

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