Breaking a Capacitor into Combinations

Q.

What is the equivalent capacitance of three capacitors?

Q. Separation between the plates of a parallel plate capacitor is and the area of each plate is A. When a slab of material of dielectric constant and thickness t(t<d) is introduced between the plates, its capacitance becomes

1. 

2. 

3. 

4. 

Q. Two thin dielectric slabs of dielectric constants $K_1$ and $K_2(K_1<K_2)$ are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $\text{P}$ is correctly shown by

1. 
2. 
3. 
4. 

Q.

A capacitor of capacitance $C_1=1.0\mu F$ can with stand a maximum voltage $V_1=6.0kV$. Anothercapacitor of capacitance $C_2=2.0\mu F$ can withstand a maximum voltage $V_2=4.0kV.$ If the capacitors are connected in series, the combination can withstand a maximum voltage of

1. 10 kV
   

2. 9 kV
   

3. 8 kV
   

4. 6 kV

Q. Capacitance of a capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates (d = separation between the plates). Then the dielectric constant of the slab is

Q. Three capacitors having capacitance $20\mu \text{F},30\mu \text{F}$ and $40\mu \text{F}$ are connected in series with a $12\text{V}$ battery. The amount of work done by the battery in charging the combination will be:

1. $2.25\times {10}^{-3}\text{J}$
2. $1.33\times {10}^{-3}\text{J}$
3. $1.5\times {10}^{-3}\text{J}$
4. $3\times {10}^{-3}\text{J}$

Q.

A capacitor of capacitance $C_1=1.0\mu F$ can with stand a maximum voltage $V_1=6.0kV$. Anothercapacitor of capacitance $C_2=2.0\mu F$ can withstand a maximum voltage $V_2=4.0kV.$ If the capacitors are connected in series, the combination can withstand a maximum voltage of

1. 10 kV
   

2. 9 kV
   

3. 8 kV
   

4. 6 kV

Q. Three dielectrics of thickness $t_1,t_2,t_3$ and dielectric constants $k_1,k_2,k_3$ respectively are placed in between the plates of a
capacitor as shown. What is the new capacitance?

1. $\frac{{\epsilon }_{0}A}{d}$
2. $\frac{{\epsilon }_0{k}_1{k}_2{k}_{3}A}{d-(t_1+t_2+t_3)}$
3. $\frac{{\epsilon }_{0}A}{d-(t_1+t_2+t_3)+\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}}$
4. None of these

Q. The capacitance of a parallel plate air condenser is 2 $\mu$ F. A dielectric of constant 4 is introduced as shown in the figure to fill half the space between the parallel plates. Its capacitance now will be?

1. $20\mu F$
2. $3.2\mu F$
3. $5.0\mu F$
4. $6.4\mu F$

Q. A parallel plate condenser initially has air as a medium between the plates. If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity will be ___

1. $33.3\%$
2. $66.7\%$
3. $50\%$
4. $75\%$