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.

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

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

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

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

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Can u verify my solution??

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Wait a minute: does the question say that

permittivity of wateris \(\epsilon\) or that therelativepermittivity of water is \(\epsilon\)?Log in to reply

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