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lorenz0

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- Homework Statement
- Three spherical thin and hollow conductors ##C_1, C_2, C_3,## have radii respectively ##R_1, R_2, R_3.## A charge ##q_1## is put on ##C_1,## a charge ##q_2## is put on ##C_2,## and a charge ##q_3## on ##C_3.##

Find: (a) the electric field ##E## at a point ##P## at distance ##d## from ##R_3;## (b) The potential difference ##V_3 -V_1## between the conductors ##C_3## and ##C_1;## (c) Now, the conductors ##C_1## and ##C_2## are joined by a conducting cable. Find the variation in electrostatic energy energy ##\Delta U_e.##

- Relevant Equations
- ##V_3-V_1=\int_{R_3}^{R_1}\vec{E}\cdot d\vec{l},\ U=\frac{1}{2}\int \sigma V da##

(a) Using Gauss's Law ##E_P=\frac{q_1+q_2+q_3}{4\pi\varepsilon_0(R_1+R_2+R_3+d)^2};(b) V_3-V_1=\int_{R_3}^{R_2}\frac{q_1+q_2}{4\pi\varepsilon_0 r^2}dr+\int_{R_2}^{R_1}\frac{q_1}{4\pi\varepsilon_0 r^2}dr=\frac{q_2}{4\pi\varepsilon_0}\left(\frac{1}{R_3}-\frac{1}{R_2}\right).##

(c) ##U_i=\frac{1}{8\pi\varepsilon_0} \left(\frac{q_1^2}{R_1}+\frac{q_2^2}{R_2}+\frac{q_3^2}{R_3}\right)## and when ##C_1## and ##C_2## are connected we have that ##q_{1f}+q_{2f}=Q##, where ##Q:=q_1+q_2## and ##V_1=V_2\Leftrightarrow \frac{q_1}{4\pi\varepsilon_0 R_1}=\frac{q_2}{4\pi\varepsilon_0 R_2}## which together imply that ##q_{1f}=\frac{R_1}{R_1+R_2}Q,\ q_{2f}=\frac{R_2}{R_1+R_2}Q## so ##U_f-U_i=\frac{1}{4\pi\varepsilon_0 R_1}\left(q_{1f}^2-q_1^2\right)+\frac{1}{4\pi\varepsilon_0 R_2}\left(q_{2f}^2-q_2^2\right)##

(c) ##U_i=\frac{1}{8\pi\varepsilon_0} \left(\frac{q_1^2}{R_1}+\frac{q_2^2}{R_2}+\frac{q_3^2}{R_3}\right)## and when ##C_1## and ##C_2## are connected we have that ##q_{1f}+q_{2f}=Q##, where ##Q:=q_1+q_2## and ##V_1=V_2\Leftrightarrow \frac{q_1}{4\pi\varepsilon_0 R_1}=\frac{q_2}{4\pi\varepsilon_0 R_2}## which together imply that ##q_{1f}=\frac{R_1}{R_1+R_2}Q,\ q_{2f}=\frac{R_2}{R_1+R_2}Q## so ##U_f-U_i=\frac{1}{4\pi\varepsilon_0 R_1}\left(q_{1f}^2-q_1^2\right)+\frac{1}{4\pi\varepsilon_0 R_2}\left(q_{2f}^2-q_2^2\right)##

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