Fixed potentials

Electricity and Magnetism Level 4

Two hemispheres of radius R = 30 cm R=30~\text{cm} are held at constant potentials V 1 = 3 V V_{1}=3~\text{V} and V 2 = 9 V V_{2}=9~\text{V} . Determine the electric potential in volts at a point P with coordinates ( x , y , z ) = ( 0 , 0 , 3 R ) (x,y,z)=(0,0, 3R) .


The answer is 2.

David Mattingly by David Mattingly

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1 solution

Albert Han
Dec 31, 2013

Based on the picture, we can assume that those two charged hemispheres are hollow.

Then, let's first consider when the whole sphere had a same voltage called V V . then, by using the shell theorem, we can calculate the voltage at point P P . We can say that, k Q R = V k \frac {Q} {R} = V and k Q 3 R = V 3 k \frac {Q} {3R} = \frac {V} {3} where Q Q equals to the charge on the spherical conductor.

Then , due to symmetry, the voltage the left hemisphere gives to the point is equal to the right one, and since the voltage is a scalar value, we can divide V 3 \frac {V} {3} in half to find the voltage given by one hemisphere. Thus, the voltage at point P when there is a hemisphere of voltage V V will be V 6 \frac {V} {6} .

Then, to find the voltage at point P P in such hemisphere, we can find two separate voltages from hemisphere of 3 V 3V and 9 V 9V and add two. Then, 3 6 + 9 6 = 12 6 = 2 \frac {3} {6} +\frac {9} {6} = \frac {12} {6} = \boxed {2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I tried doing like this whats wrong in my method. Let the charges on the two be Q 1 , Q 2 Q_{1},Q_{2}

At any point outside the hemispheres on the symmetrical axis the radial component at a distance r from the centre of electric field is given by. E z = K ( Q 1 + Q 2 ) r 2 E_{z}=\frac{K(Q_{1}+Q_{2})}{r^{2}}

Therefore the potential difference can be given by

∫ R 3 R ( − E z d r ) = − 2 K ( Q 1 + Q 2 ) 3 R = V 3 R − V R \int_{R}^{3R}(-E_{z}dr)=-\frac{2K(Q_{1}+Q_{2})}{3R}=V_{3R}-V_{R}

If we consider only a single hemisphere at a time say Q_{1} the potential due to a single hemisphere at its surface (near the point on the symmetrical axis at a distance R from the centre) = V 1 = K Q 1 R = 9 V_{1}=\frac{KQ_{1}}{R}=9

similar for the other hemisphere. V 2 = K Q 2 R = 3 V_{2}=\frac{KQ_{2}}{R}=3

By superposition. Thus the potential V R = ( V 1 + V 2 ) = 12 V_{R}=(V_{1}+V_{2})=12

Thus we get V 3 R = 12 3 = 4 V_{3R}=\frac{12}{3}=4

Milun Moghe - 7 years, 3 months ago

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