22P3Q04

4a)

If the spheres were oppositely charged, they would produce potentials of opposite signs. This would have resulted in a point between the spheres where the total potential is zero.

Both spheres must be positively charged since the potentials in the graph are all positive.

4b)

At \displaystyle x=0.15\text{ m}, \displaystyle E=-\frac{dV}{dr}=-\frac{620-80}{0.15-0.28}=4150\text{ N }{{\text{C}}^{-1}}

At \displaystyle x=0.40\text{ m}, \displaystyle E=-\frac{dV}{dr}=-\frac{360-100}{0.18-0.66}=542\text{ N }{{\text{C}}^{-1}}

At \displaystyle x=0.68\text{ m}, \displaystyle E=-\frac{dV}{dr}=0\text{ N }{{\text{C}}^{-1}}

At \displaystyle x=0.80\text{ m}, \displaystyle E=-\frac{dV}{dr}=-\frac{330-120}{0.90-0.72}=-1170\text{ N }{{\text{C}}^{-1}}

4ci)

\displaystyle \begin{aligned}   (V&=k\frac{Q}{r}) \\  V&=(8.99\times {{10}^{9}})\frac{1.0\times {{10}^{-8}}}{0.15} \\  &=599\text{ V} \end{aligned}

4cii)

The actual electric potential on A’s surface is the summation of the electric potential due to both sphere A and B.

COMMENT: It is true that the charges on sphere A and B may not be uniformly distributed thanks to their mutual repulsion. However, this would have led to the charges moving further away from the \displaystyle x=0.15\text{ m}point in the graph. This would only result in the value of V to be lower, not higher than what’s calculated.

COMMENT: Assuming the potential due to sphere B to be about 100 V at \displaystyle x=0.80\text{ m}as the graph suggests, it would contribute about \displaystyle \frac{0.90-0.80}{0.90-0.15}\times 100=\frac{0.10}{0.75}\times 100=13\text{ V}to the potential at \displaystyle x=0.15\text{ m}.

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