<h2>
Answer:</h2>
455pF
<h2>
Explanation:</h2>
Given;
Turning dial from 0° to 180° changes;
capacitance from 100 to 350pF
Also, we know that the capacitance (C) of a capacitor is related to the charge(Q) on and voltage (V) across the capacitor as follows;]
Q = CV --------------------------(i)
Now, since at 180° the capacitor is connected to 130V, this means that the capacitance of the capacitor at that point is 350pF.
i.e
at V = 130V, C = 350pF = 3.50 x 10⁻¹²F
Substitute these values into equation (i) as follows;
Q = 3.50 x 10⁻¹² x 130
Q = 455 x 10⁻¹²C
Q = 455pF
Therefore, the charge of the capacitor at 180°, 350pF is 455pF.
But, since the charge on a capacitor remains unless it is discharged, the charge of the capacitor even at 0° at a disconnected battery source remains 455pF.
