If E be the electric field inside a parallel plate capacitor due to Q and -Q charges on the two plates, then electrostatic force on plate having charge -Q due to the plate having charge +Q will be

1. -QE

2. $\frac{-QE}{2}$

3. QE

4. $\frac{-QE}{4}$

Capacitors $C_1=10\mu F<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>C_2=30\mu F$ are connected in series across a source of emf 20KV. The potential difference across $C_1$ will be

1. 5 KV

2. 15 KV

3. 10 KV

4. 20 KV

A parallel plate capacitor is made by stacking n equally spaced plates connected alternately. If the capacitance between any two plates is C then the resultant capacitance is

1. C

2. nC

3. (n – 1)C

4. (n + 1)C

Four plates of equal area A are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity is

1. $\frac{2{\epsilon }_{0}A}{d}$

2. $\frac{3{\epsilon }_{0}A}{d}$

3. $\frac{4{\epsilon }_{0}A}{d}$

4. $\frac{{\epsilon }_{0}A}{d}$

Three capacitors are connected to D.C. source of 100 volts shown in the adjoining figure. If the charge accumulated on plates of C₁, C₂ and C₃ are $q_a,q_b,q_c,q_d.q_e$ and q_f respectively, then 

1. $q_b+q_d+q_f=\frac{100}{9}C$

2. $q_b+q_d+q_f=0$

3. $q_a+q_c+q_e=50C$

4. $q_b=q_d=q_f$