A capacitor of plate area 0.01 m2 and plate separation 0.05 mm is charged by a battery of potential 10 V then the charge on positive plate will beChoose answer: 0.18 μC 0.17 nC 8.8 nC 17.7 nC
Question
A capacitor of plate area 0.01 m2 and plate separation 0.05 mm is charged by a battery of potential 10 V then the charge on positive plate will beChoose answer: 0.18 μC 0.17 nC 8.8 nC 17.7 nC
Solution
To solve this problem, we need to use the formula for the capacitance of a parallel plate capacitor, which is:
C = ε0 * (A/d)
where: C is the capacitance, ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of one of the plates (0.01 m^2), and d is the separation between the plates (0.05 mm = 0.05 x 10^-3 m).
Substituting the given values into the formula, we get:
C = 8.85 x 10^-12 F/m * (0.01 m^2 / 0.05 x 10^-3 m) = 1.77 x 10^-12 F
The charge (Q) on a capacitor is given by the formula Q = CV, where V is the voltage across the capacitor. Substituting the values we have:
Q = 1.77 x 10^-12 F * 10 V = 1.77 x 10^-11 C
Converting this to nC (nanoCoulombs), we get:
Q = 1.77 x 10^-11 C * (1 x 10^9 nC/1 C) = 17.7 nC
So, the charge on the positive plate of the capacitor is 17.7 nC.
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