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PHY7C FNT9_alpha

# PHY7C FNT9_alpha - FNT#9 Electric fields and voltages...

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FNT #9: Electric fields and voltages Damien Martin November 12, 2005 Question 1: Electric field and electric voltage The electric field E and the voltage V at a point r = R are related by E ( r = R ) = vextendsingle vextendsingle vextendsingle vextendsingle Δ V Δ R vextendsingle vextendsingle vextendsingle vextendsingle This tells us that the electric field is related to how quickly the voltage changes. Here the “Δ” means we are looking at finite differences in the voltage and distance, and doing a rise over run. A more elegant and precise way of doing it is to look at the tangent at the same point: E ( r = R ) = vextendsingle vextendsingle vextendsingle vextendsingle d V d r vextendsingle vextendsingle vextendsingle vextendsingle Written in this form, we are reminded that the functional form of E can be obtained from the functional form of V via a derivative. Given that the field of a point change E is given by E = k | Q | r 3 find the correct expression for the Voltage from a charge. Solution This question is a little bit tricky. To start with, we have that the electric field is the negative slope of the electric field: E = - d V d r . This should be given in the problem, but it is in your notes. 1

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The rest of this problem is realising that to “undo” a derivative we need to integrate. That is V = integraldisplay d V d r d r. This is the fundamental theorem of calculus . If you have trouble remembering it, the easiest way is to think of the d r on the “denominator” and the d r in the “numerator” cancelling: integraldisplay d V a26 a26 d r a26 a26 d r = integraldisplay d V = V. To finish the problem, we need to actually do the integral: V = integraldisplay d V d r d r = integraldisplay ( - E ) d r = - kQ integraldisplay 1 r 2 d r = + kQ r Question 2: An α particle An α
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PHY7C FNT9_alpha - FNT#9 Electric fields and voltages...

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