Figs. 1 and 2 show a dielectric slab being inserted between the plates of one of
ID: 1546937 • 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 1 ____ when the dielectric is inserted.
B) In Fig. 2, the charge on capacitor 1 ____ when the dielectric is inserted.
C) In Fig. 2, the capacitance of capacitor 1 ____ when the dielectric is inserted.
D) In Fig. 2, the potential energy stored in capacitor 1 ____ when the dielectric is inserted.
Figure 1 #1 #2 Figure 2 #1 #2Explanation / Answer
C1 =C2 = C
in fig-2 when dilectric is inserted then its capacitance will be, C2 = k C
charge Q = CV
V= Q/ C
ie V(potential) is inversly proportional to Capacitance.
V2 ( potential on C2 ) = 1/ K C ( Q = constant)
when dilectric slab is being inserted ,V2 decreases ( since Capacitance increases).
In parallel combination potential ,V= V1 = V2
since V2 decreses so V1(potential on C1) also decreses.
A) In Fig. 2, the potential difference between the plates of capacitor 1 _D___ when the dielectric is inserted.
B) in parallel combination ,total charge Q = Q1 + Q2
since Q2 = KC*V ( from Q= CV)
ie Q2 increses when dilectric slab is inserted.
So Q1 will decrease to kept Q same.
In Fig. 2, the charge on capacitor 1 __D__ when the dielectric is inserted.
C) C1 = C ( same as previous one)
In Fig. 2, the capacitance of capacitor 1 __S__ when the dielectric is inserted.
D) Total energy U= U1 +U2
U2 ( energy of C2) = (1/2)*K*C*V2 ( From U= (1/2)*C*V2 )
since U2 increses ,So U1 will decrease to kept U same.
In Fig. 2, the potential energy stored in capacitor 1 __D__ when the dielectric is inserted.
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