# Problem № 8 (15 points)

50 identical metal spheres with a radius of $R=1$ mm were connected by equal conducting segments into a chain, with the length of each connecting wire segment $l=30$ cm being much greater than the radius of the sphere. The resulting structure was then placed in a uniform electric field of known intensity $E=100 V/m$. The spheres are arranged in a straight line parallel to the direction of the electric field. What magnitude of charges are induced on the outermost spheres in the chain.
Answer: $8.17 \cdot 10^{-11} C$

## Solution and Grading Criteria:

Since the spheres are far apart from each other, they can be considered isolated, i.e., the mutual influence on each other can be neglected. The potential of the field created by a sphere is:

$\varphi=k \frac{q}{R}$

The charges of the outermost spheres are equal in magnitude but opposite in sign.

(2 points) A conductor in an electric field is an equipotential surface, so the potential difference between the outermost spheres should be zero, and we get:

$0=\Delta \varphi_{B}+\left(\varphi_{N}-\varphi_{1}\right)=\Delta \varphi_{B}+2 \varphi_{1}=\Delta \varphi_{B}+2 k \frac{q}{R}$

where $\Delta \varphi_{B}=E(N-1) l$ is the potential difference created by the external field. (3 points)

We obtain: $E(N-1) l=2 k \frac{q}{R}$

Final result: $q=\frac{E(N-1) l R}{2 k}=\frac{100 \cdot 49 \cdot 0.3 \cdot 0.001}{2 \cdot 9 \cdot 10^{9}}=8.17 \cdot 10^{-11}$ C

Multiprofile Engineering Olympiad "Star" in Natural Sciences

## Final Stage

2016-2017 academic year

Tasks, answers, and grading criteria

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