CapacitanceEasy
Question
Two identical metal plates are given positive charge Q1 and Q2 ( < Q1) respectively. If they are now brought close together to from a parallel plates capictors with capacitance C, the potential difference between them is
Options
A.(Q1 + Q2)/ 2C
B.(Q1 + Q2)/ C
C.(Q1 - Q2)/ C
D.(Q1 - Q2)/ 2C
Solution
Electric field within the plates
E = E1 - E2
∴ Potential difference between the plates
AA - B = Ed
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