Answer: $2.75 \cdot 10^{-9}$ Cl

## Solution and Grading Criteria:

Since the balls 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 ball is:

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

The charges of the outermost balls are equal in magnitude but different in sign. A conductor in an electric field is an equipotential surface, so the potential difference between the outermost balls 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: $q=\frac{E(N-1) l R}{2 k}=\frac{1000 \cdot 99 \cdot 0.5 \cdot 0.001}{2 \cdot 9 \cdot 10^{9}}=2.75 \cdot 10^{-9} \mathrm{Cl}$
# Problem № 8 (15 points)

100 identical metal balls with a radius of $R=1$ mm were connected by equal conducting segments into a chain, with the length of each connecting wire segment $l=50$ cm being much greater than the radius of the ball. The resulting structure was then placed in a uniform electric field of known intensity $E=1000 V / m$. The balls are arranged in a straight line parallel to the direction of the electric field. What magnitude of charges are induced on the outermost balls in the chain.