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 $C_1=10\mu F and C_2=30\mu F$ 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) $q_b=q_d=q_f$