Figs. 1 and 2 show a dielectric slab being inserted between the plates of one of
ID: 1542378 • Letter: F
Question
Figs. 1 and 2 show a dielectric slab being inserted between the plates of one of two identical capacitors, capacitor 2. Select the correct answer to each of the statements below (enter I for "increases', D for "decreases', or S for "stays the same'). A. In Fig. 2, the potential difference between the plates of capacitor 2__when the dielectric is inserted. B. In Fig. 2, the charge on capacitor 1___when the dielectric is inserted. C. In Fig. 1. the capacitance of capacitor 2___when the dielectric is inserted. D. In Fig. 2, the potential energy stored in capacitor 2___when the dielectric is inserted.Explanation / Answer
A.) In figure 2, the potential difference between the plates of capacitor 2 stays the same when the dielectric is inserted since the plates are still across Va and Vb even after the dielectric in inserted.
B.) The charge on capacitor is given by Q = CV Here the capacitance C is a function of its dimensions and hence remains constant. Also, the potential difference across capacitor 1 remains the same since the plates are still connected across Va and Vb . So, since C and V do not change, Q also will not change and stays the same.
C.) Capacitance is given by C = kA/d where k is the dielectric constant.
So, as we see from the formula, the capacitance is directly proportional to the dielectric constant and hence increases as we insert the dielectric.
D.) Potential energy stored in a parallel plate capacitor is given by U = 0.5 CV2
Since the V here remains constant as discussed in part A, and also since the C here increases because the capacitance indirectly proportional to the dielectric constant; the potential energy will increase with the increase in C as the dielectric is inserted.
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