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Electrical Potential Question

Can someone please explain to me how you solve question 15 on this paper? Thanks

http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF
Original post by Dentist95
Can someone please explain to me how you solve question 15 on this paper? Thanks

http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF


The electric potential ϕ\phi is the work done in moving a unit charge to a location, from infinity, in an electric field. For a charge qq:

ϕ=qkr\phi = \frac{q}{kr}

where kk is a constant and rr is the distance from the unit charge to qq.

Here we have to consider the electric potential due to two charges qq. It turns out that we simply add the potentials for each i.e. they obey the principle of superposition (if one was -ve, we'd have to take account of that though via setting the charge to q-q)

We are interested in the line from P to Q. Let PU=xPU =x be the distance from P to the location of the unit charge. Then UQ=4xUQ = 4-x and the total potential as a function of xx due to the two charges is:

ϕ(x)=ϕP+ϕQ=qkx+qk(4x)=qk(1x+14x)\phi(x) = \phi_P + \phi_Q = \frac{q}{kx} + \frac{q}{k(4-x)} = \frac{q}{k}(\frac{1}{x}+\frac{1}{4-x})

You know that at the central point, x=2x=2 and ϕ(2)=25\phi(2) = 25. You should be able to do the rest.

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