# 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
2 years, 11 months ago

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

- 2 years, 11 months ago

Can u verify my solution??

- 2 years, 11 months ago

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

- 2 years, 11 months ago

But your method is fine.

- 2 years, 11 months ago

Thanks for reviewing it!

- 2 years, 11 months ago

Relative permittivity

- 2 years, 11 months ago

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

Staff - 2 years, 11 months ago

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

- 2 years, 11 months ago

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

Staff - 2 years, 11 months ago

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?

- 1 month ago

Thanks a lot sir!

- 2 years, 11 months ago