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.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Josh Silverman @Ronak Agarwal @Mvs Saketh @Deepanshu Gupta @Calvin Lin @Best of Electricity and Magnetism @Nishant Rai @Atri Dutta @NILAY PANDE and others please help.

Aditya Kumar - 5 years, 9 months ago

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

Josh Silverman Staff - 5 years, 9 months ago

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

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

@Josh Silverman – Thanks a lot sir!

@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 - 2 years, 12 months ago

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

Raj Magesh - 5 years, 9 months ago

Can u verify my solution??

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

@Raj Magesh – Relative permittivity

@Raj Magesh – But your method is fine.

@Raj Magesh – Thanks for reviewing it!

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$