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AS Physics exam question: Have I answered this correctly? watch

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    Q) A square sheet of carbon-reinforced plastic, measuring 90 mm x 90 mm and 1.1 mm thick, has its two large surfaces coated with a highly conducting metal film. When a p.d. of 210 V is applied between the metal films a current of 1.4mA passes through the plastic sheet. Calculate the resistivity of the plastic.

    So I know that:

    P = \frac{RA}{L}

    So what I did first was work out resistance.

    Since \frac{V}{I} = R then \frac{210V}{(1.4 \times 10^-3)A} = 150,000 Ohms

    Then I worked out the cross-sectional area:

    1.1 mm \times 90 mm = 99 mm^2

    And finally the length:

    90 mm \times 4 = 360 mm

    So placing that into the resistivity equation:

    P = \frac{150,000 \times 99}{360} = 41250 Ohm Metres

    I have gone wrong somewhere and it's probably when I did the area or maybe the length but I need that pointing out because I'm no good at spotting these things...
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    cross sectional area is not 1.1mm x 90mm, 1.1 is the length
    No idea what the 90mm x 4 was, that's the perimeter. the area is 90mm x 90mm.
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    (Original post by KissMyArtichoke)
    cross sectional area is not 1.1mm x 90mm, 1.1 is the length
    No idea what the 90mm x 4 was, that's the perimeter. the area is 90mm x 90mm.
    But I thought 1.1 was the width? I understand the 90x90 as the area because I kinda had a feeling I had done something wrong before, but why is 1.1 the length...
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    you have to think of it as a wire. The current flows from one surface to the other. The distance between them is always the length. Take a wire, the length is obvious. Now slice a bit off the end. The length is now smaller than the diameter, but it's still the length. The only difference here is the wire is square. It's just a really, really short wire.
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    (Original post by KissMyArtichoke)
    you have to think of it as a wire. The current flows from one surface to the other. The distance between them is always the length. Take a wire, the length is obvious. Now slice a bit off the end. The length is now smaller than the diameter, but it's still the length. The only difference here is the wire is square. It's just a really, really short wire.
    Hmm I can do that; the problem is I can only visualise it as a really fat wire.

    Ah well just gotta accept it :P
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    Made you a pic. The only difference in the second wire is that it's much shorter, the length is tiny. You still treat it the same way
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    (Original post by KissMyArtichoke)
    Made you a pic. The only difference in the second wire is that it's much shorter, the length is tiny. You still treat it the same way
    Aah thanks alot mate that helps me visualise it soo much better
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    (Original post by KissMyArtichoke)
    Made you a pic. The only difference in the second wire is that it's much shorter, the length is tiny. You still treat it the same way
    One more question; do you have to convert the measurements mm into anything?
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    The top one is the one in the question the bottom one is a wire. (The length in the top is 1.1mm)

    You can see how they are similar? only difference is that a wire's resistivity is much lower.
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    (Original post by Jambre)


    The top one is the one in the question the bottom one is a wire. (The length in the top is 1.1mm)

    You can see how they are similar? only difference is that a wire's resistivity is much lower.
    Yeah I see, so as my final answer I got:

    P = \frac{150,000 \times 8.1}{1.1\times10^-3} = 1104545455

    Should I put that in standard form, or is that answer actually wrong?
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    (Original post by Dededex)
    Q) A square sheet of carbon-reinforced plastic, measuring 90 mm x 90 mm and 1.1 mm thick, has its two large surfaces coated with a highly conducting metal film. When a p.d. of 210 V is applied between the metal films a current of 1.4mA passes through the plastic sheet. Calculate the resistivity of the plastic.

    So I know that:

    P = \frac{RA}{L}

    So what I did first was work out resistance.

    Since \frac{V}{I} = R then \frac{210V}{(1.4 \times 10^-3)A} = 150,000 Ohms

    Then I worked out the cross-sectional area:

    1.1 mm \times 90 mm = 99 mm^2

    And finally the length:

    90 mm \times 4 = 360 mm

    So placing that into the resistivity equation:

    P = \frac{150,000 \times 99}{360} = 41250 Ohm Metres

    I have gone wrong somewhere and it's probably when I did the area or maybe the length but I need that pointing out because I'm no good at spotting these things...
    Just so that you are aware, 'Resistivity' is given the symbol 'rho' (i dont have it on my keyboard to be able to type it in this message, but if you type 'rho' into google images/wikipedia you will see the greek letter i mean, it looks like a lower case p but more rounded) which is a greek letter- you have given it a Captial 'P' which is incorrect, at least according to any textbook ive ever seen on the subject.
    i just thought i better point this out incase you lose a mark in exams for using the wrong symbol in a question.
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    (Original post by bloomblaze)
    Just so that you are aware, 'Resistivity' is given the symbol 'rho' (i dont have it on my keyboard to be able to type it in this message, but if you type 'rho' into google images/wikipedia you will see the greek letter i mean, it looks like a lower case p but more rounded) which is a greek letter- you have given it a Captial 'P' which is incorrect, at least according to any textbook ive ever seen on the subject.
    i just thought i better point this out incase you lose a mark in exams for using the wrong symbol in a question.
    Thanks, don't think i'd get marked down though unless maybe it asked me for the equation which nowadays is in the front of the book so I doubt they would.

    But thanks for letting me know.
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    (Original post by Dededex)
    Yeah I see, so as my final answer I got:

    P = \frac{150,000 \times 8.1}{1.1\times10^-3} = 1104545455

    Should I put that in standard form, or is that answer actually wrong?
    You have converted millimetres incorrectly for the area. According to what you've done a 90mm x 90mm square has a cross sectional area of 8.1m!! does that sound reasonable? Because it is squared you have to times it by (10-3)^2 or divide by a million if you like, rather than a thousand. If that confuses you, you can convert all the mm into m before you do the calculations and it will work just as fine.

    Also the question only quotes the figures to two significant figures, so your answer should be to that too. Otherwise, your answer will be more precise than the question allows it to be!
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    (Original post by Dededex)
    Q) A square sheet of carbon-reinforced plastic, measuring 90 mm x 90 mm and 1.1 mm thick, has its two large surfaces coated with a highly conducting metal film. When a p.d. of 210 V is applied between the metal films a current of 1.4mA passes through the plastic sheet. Calculate the resistivity of the plastic.

    So I know that:

    P = \frac{RA}{L}

    So what I did first was work out resistance.

    Since \frac{V}{I} = R then \frac{210V}{(1.4 \times 10^-3)A} = 150,000 Ohms

    Then I worked out the cross-sectional area:

    1.1 mm \times 90 mm = 99 mm^2

    And finally the length:

    90 mm \times 4 = 360 mm

    So placing that into the resistivity equation:

    P = \frac{150,000 \times 99}{360} = 41250 Ohm Metres

    I have gone wrong somewhere and it's probably when I did the area or maybe the length but I need that pointing out because I'm no good at spotting these things...
    You have got a little confused here, so I'll try to help simplify.

    Resistivity equation= all good

    Resistance value of the metal sheet= all good

    Cross sectional area. You have got this a little wrong here.

    Firstly, it is a square sheet so length and width are both 90mm. Im not sure where you got the 4 from for the length?

    Cross sectional area is length x height x width. So, 90mm= 0.09 metres. 1.1mm is 0.0011 metres.

    0.09 x 0.09 x 0.0011= cross sectional area

    From what I have interpreted, the length isn't 1.1mm. Because thats the height of the sheet.

    So then putting into the resistivity equation you get:

    P = \frac{150000A}{0.09}

    Spoiler:
    Show

    Cross sectional Area= 0.09*0.09*0.0011= 8.91X10^-6

    P = \frac{150000*(8.91X10^-6)}{0.09} = 14.85


    *Prays he hasn't just embarrassed himself by getting it wrong*
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    (Original post by Jambre)
    You have converted millimetres incorrectly for the area. According to what you've done a 90mm x 90mm square has a cross sectional area of 8.1m!! does that sound reasonable? Because it is squared you have to times it by (10-3)^2 or divide by a million if you like, rather than a thousand. If that confuses you, you can convert all the mm into m before you do the calculations and it will work just as fine.

    Also the question only quotes the figures to two significant figures, so your answer should be to that too. Otherwise, your answer will be more precise than the question allows it to be!
    Aah yeah thanks alot! ofcourse when you times powers you add subtract accordingly - thanks.
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    (Original post by fishkeeper)
    You have got a little confused here, so I'll try to help simplify.

    Resistivity equation= all good

    Resistance value of the metal sheet= all good

    Cross sectional area. You have got this a little wrong here.

    Firstly, it is a square sheet so length and width are both 90mm. Im not sure where you got the 4 from for the length?

    Cross sectional area is length x height x width. So, 90mm= 0.09 metres. 1.1mm is 0.0011 metres.

    0.09 x 0.09 x 0.0011= cross sectional area

    From what I have interpreted, the length isn't 1.1mm. Because thats the height of the sheet.

    So then putting into the resistivity equation you get:

    P = \frac{150000A}{0.09}

    Spoiler:
    Show

    Cross sectional Area= 0.09*0.09*0.0011= 8.91X10^-6

    P = \frac{150000*(8.91X10^-6)}{0.09} = 14.85


    *Prays he hasn't just embarrassed himself by getting it wrong*
    You're joking? Now everyone has got different answers I don't know which one is right - some have said that the length is 1.1 but you're saying it's not?
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    Where is this question from? Its either a paper or an exam style question in your book?

    I don't see how 1.1mm can be the length. If you looked at this sheet from above, it would appear a 2D square with a width and length and you could easily work out the area by 90mm x 90mm yes?

    Well, the 1.1 is just adding a height to this 3D shape. Unless you said the sheet was upright then 1.1 cannot be the length. The length is still the same as it was in the 2D shape...the 90mm side.
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    (Original post by Jambre)
    You have converted millimetres incorrectly for the area. According to what you've done a 90mm x 90mm square has a cross sectional area of 8.1m!! does that sound reasonable? Because it is squared you have to times it by (10-3)^2 or divide by a million if you like, rather than a thousand. If that confuses you, you can convert all the mm into m before you do the calculations and it will work just as fine.

    Also the question only quotes the figures to two significant figures, so your answer should be to that too. Otherwise, your answer will be more precise than the question allows it to be!
    So wait if I said:

    p = \frac{150,000 \times (8.1 \times 10^-3)}{(1.1 \times 10^-3)}

    That would be correct right?
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    (Original post by fishkeeper)
    Where is this question from? Its either a paper or an exam style question in your book?
    It's from a paper - but my teacher has kind of done a photocopied compilation of exam questions. It's "Jan 07/PHA3/W".
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    (Original post by Dededex)
    So wait if I said:

    p = \frac{150,000 \times (8.1 \times 10^-3)}{(1.1 \times 10^-3)}

    That would be correct right?
    No, you have still done the 90 x 90 calculation before factoring the change in units. If you write (90 x 10^-3) x (90 x 10^-3) in place of the (8.1 x 10^-3) then it would be correct.

    Edit: ignore what fishkeeper us saying, it is wrong. The "length" in terms of the resistivity is the distance between the points you are measuring and is perpendicular to the surface area. In this case it is the thickness of the square sheet, whereas the surface area is given by the area of the square face. If you cut anywhere along the the thickness, it will still have the same surface area.
 
 
 
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