Two electric charges e_0=100 CGSE and e_1=200 CGSE lie
Two electric charges $e_0=100$ CGSE and $e_1=200$ CGSE lie on the $x$-axis at $x_0=0$ and $x_1=1$ cm. How much work is done by the electric field if the second charge moves to point $x_2=10$ cm?
Explanation of Method
In electrostatic (CGS-ESU) units, two point charges separated by distance $r$ have potential energy $U=e_0e_1/r$. The work done by the field as the configuration changes equals the decrease in potential energy, $U_{\text{initial}}-U_{\text{final}}$; equivalently, it is the integral of the Coulomb force $F(r)=e_0e_1/r^2$ over the distance moved.
Step 1: Potential energy before and after
Initial separation $r_1=1$ cm, final separation $r_2=10$ cm:
\[U_1=\frac{e_0e_1}{r_1}=\frac{100\times200}{1}=20000\ \text{erg},\qquad U_2=\frac{e_0e_1}{r_2}=\frac{100\times200}{10}=2000\ \text{erg}.\]
Step 2: Work done by the field
\[W=U_1-U_2=20000-2000=18000\ \text{erg}.\]
\[\boxed{W=18000\ \text{erg}}\]
Step 3: Cross-check by direct integration of the force
Since both charges are positive, they repel with force $F(r)=e_0e_1/r^2$ along the direction of increasing $r$:
\[W=\int_1^{10}\frac{e_0e_1}{r^2}\,dr=20000\int_1^{10}\frac{dr}{r^2}=20000\left[-\frac1r\right]_1^{10}=20000\left(1-\frac1{10}\right)=18000\ \text{erg},\]
matching Step 2 exactly.
Step 4: Sign and interpretation
Because the charges are both positive (repulsive), moving them farther apart is the direction the field itself pushes, so the field does positive work $(+18000\ \text{erg})$ and releases energy; an external agent restraining the motion would instead do work $-18000$ erg. In SI-adjacent units, $18000\ \text{erg}=1.8\times10^{-3}\ \text{J}=1.8\ \text{mJ}$.
Result
Final answer:
\[\boxed{W=18000\ \text{erg}=1.8\ \text{mJ}\ \text{(done by the field, since the charges repel and move apart)}}\]