A) \[\frac{{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}}{2C}\]
B) \[\frac{{{Q}_{2}}+{{Q}_{3}}}{2C}\]
C) \[\frac{{{Q}_{2}}-{{Q}_{3}}}{2C}\]
D) \[\frac{{{Q}_{1}}+{{Q}_{4}}}{2C}\]
Correct Answer: C
Solution :
Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, \[{{Q}_{2}}=-{{Q}_{3}}\]. The charge on a capacitor means the charge on the inner surface of the positive plate (here it is \[{{Q}_{2}}\]) Potential difference between the plates \[=\frac{charge}{capacitance}\]\[=\frac{{{Q}_{2}}}{C}=\frac{2{{Q}_{2}}}{2C}\]
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