Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hey, I'm really stuck on this question and not sure where to start or how to work it. If someone could explain how to work it, that'd be great.

    For a charge of +4μC and a charge of -6μC placed 150mm apart.

    a) Find the point at which the total potential is zero
    b) Calculate the field strength at this point
    Offline

    10
    You start a) by using the formula for potential at a point a distance r from a charge Q.
    The point in question is somewhere between the two charges on the line joining them.
    If you let this point be, say, a distance r from the 4uC change, then put r into the formula for potential; then that point must be a distance 150-r from the other change.
    Put 150-r into the formula for potential due to the -6uC charge.
    Finally, for the total potential to be zero, these two values must be equal and opposite. One plus and one minus but of equal magnitude. Solve for the value of r.
    For part b) once you know the distances involved, find the field at the point due to each charge using the formula, and add the two fields together as vectors.
    NB. Watch the units. The distance is in mm at the moment.
    Do you know the formulas for potential (depends on 1/r) and field (depends on 1/r^2)?
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks, I get what you mean now. Yeah I know the formulas. Just I wasn't sure where to start. Thank you
    • Thread Starter
    Offline

    0
    ReputationRep:
    Shame I totally fail at rearranging equations.
    Offline

    0
    ReputationRep:
    Hmmm I have answers, no idea whether they are right though.. I just did them.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Haha, what are they? I am using the formula for electric potential V= KQ/R where K is (1/4 x pi x epsilon zero ) and Q is charge and R is distance from the point. But I'm still totally lost, I tried to rearrange it.

    also, noticed you applying for medicine. I am too, I also applied for Leeds and Manchester. Leeds rejected me, and Manchester I'm on hold...
    Offline

    0
    ReputationRep:
    60mm and 90mm are the distances and i get the electric field strength to be 16.6MVm^-1

    Almost the same as me in the medicine department then :P
    • Thread Starter
    Offline

    0
    ReputationRep:
    Would it just not be infinity, I'm sensing this is a trick question.
    Offline

    2
    ReputationRep:
    Well normally V(\infty) \equiv 0 and I don't think they'd just make you say that :p:

    You should have an equation like this:
    Spoiler:
    Show


    V_1 = V_2 \Rightarrow k\frac{q_1}{r_1} = k\frac{q_2}{r_2} \Rightarrow \frac{4}{r} = \frac{-6}{150-r} \Rightarrow \frac{150-r}{r} = \frac{-6}{4} \Rightarrow \frac{150}{r} - 1 = -1.5 \Rightarrow r = -30

    EDIT Ummm, on second thought, I don't really know if I did that right cause apparently r is... negative... eh?
    Offline

    0
    ReputationRep:
    (Original post by trm90)
    Well normally V(\infty) \equiv 0 and I don't think they'd just make you say that :p:

    You should have an equation like this:
    Spoiler:
    Show


    V_1 = V_2 \Rightarrow k\frac{q_1}{r_1} = k\frac{q_2}{r_2} \Rightarrow \frac{4}{r} = \frac{-6}{150-r} \Rightarrow \frac{150-r}{r} = \frac{-6}{4} \Rightarrow \frac{150}{r} - 1 = -1.5 \Rightarrow r = -30

    EDIT Ummm, on second thought, I don't really know if I did that right cause apparently r is... negative... eh?
    Ive done what you've done but I think you're wrong because..

    the minus sign on the 6 disappears because V is a scalar quantity. V+V=0 but the second V will be negative so really it is V+(-V)=0 which is V-V=0. Therefore V=V so the negative disappears

    K \times \frac{Q_1}{r_1}+K \times \frac{-Q_2}{r_2}=0

    \mathrm{therefore}

    K \times \frac{Q_1}{r_1}=K \times \frac{Q_2}{r_2}

    \frac{Q_1}{r_1}=\frac{Q_2}{r_2}

    \frac{4 \times 10^6}{r_1}=\frac{6 \times 10^6}{r_2}

    r_2=(150 \times 10^-3-r_1)

    \frac{4 \times 10^6}{r_1}=\frac{6 \times 10^6}{(150 \times 10^-3-r_1)}

    \frac{(150 \times 10^-3-r_1)}{r_1}=\frac{6 \times 10^6}{4 \times 10^6}

    \frac{(150 \times 10^-3)}{r_1}-1=\frac{3}{2}

    \frac{(150 \times 10^-3)}{r_1}=\frac{5}{2}

    \frac{150 \times 10^-3 \times 2}{5}=r_1

    0.06=r_1
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 23, 2010
The home of Results and Clearing

3,928

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  2. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
  3. Bishop Grosseteste University
    All Courses Undergraduate
    Fri, 17 Aug '18
Poll
A-level students - how do you feel about your results?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.