A) \[{{q}_{1}}=4\pi {{\varepsilon }_{0}}\left( \frac{{{\phi }_{1}}-{{\phi }_{2}}}{{{a}_{2}}-{{a}_{1}}} \right){{a}_{1}}{{a}_{2}}\]
B) \[{{q}_{1}}=4\pi {{\varepsilon }_{0}}\left( \frac{{{\phi }_{1}}+{{\phi }_{2}}}{{{a}_{2}}+{{a}_{1}}} \right){{a}_{1}}{{a}_{2}}\]
C) \[{{q}_{1}}=4\pi {{\varepsilon }_{0}}\left( \frac{{{\phi }_{1}}-{{\phi }_{2}}}{{{a}_{2}}+{{a}_{1}}} \right){{a}_{1}}{{a}_{2}}\]
D) \[{{q}_{1}}=4\pi {{\varepsilon }_{0}}\left( \frac{{{\phi }_{1}}+{{\phi }_{2}}}{{{a}_{2}}-{{a}_{1}}} \right){{a}_{1}}{{a}_{2}}\]
Correct Answer: A
Solution :
[a] \[{{\phi }_{1}}=\frac{k{{q}_{1}}}{{{a}_{1}}}+\frac{k{{q}_{2}}}{{{a}_{2}}}\] \[{{\phi }_{2}}=\frac{k{{q}_{1}}}{{{a}_{2}}}+\frac{k{{q}_{2}}}{{{a}_{2}}}\] Solve to get \[{{q}_{1}}=4\pi {{\varepsilon }_{0}}\left[ \frac{{{\phi }_{1}}-{{\phi }_{2}}}{{{a}_{2}}-{{a}_{1}}} \right]{{a}_{1}}{{a}_{2}}\]You need to login to perform this action.
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