Electrostatic P...

The total capacitance between points A and B in the arranged shown below is

Four metallic plates each with surface area '$$A$$' and inter plate separation '$$d$$' are placed as shown. Alternative plates are connected to point $$A$$ and $$B$$. The equivalent capacitance of the system is

For section $$AB$$ of a circuit, $$C_1 = 1 \mu F,$$ $$C_2 = 2 \mu F,$$ $$E = 10 V,$$ and the potential difference $$V_A - V_B = -10 V$$. Charge on capacitor $$C_1$$ is

The equivalent capacitance between A and B is

Five plates (same plate area for each) are placed as shown
in figure.Capacitance of each parallel plate capacitor is C. Then find $${C_{AB}}$$.

A combination arrangement of the capacitors is shown in the figure $$C_1 = 3\mu F, C_2 = 6\mu F$$ and $$C_3 = 2\mu F$$ then eqivalent capactance between 'a' and 'b' is:
 

In the circuit shown in figure, point C is connected to A to charge the capacitor $$4    \mu F$$ capacitor thereby changes by a factor:  

The equivalent capacitance in the figure shown between indicated points is -

In a region of uniform field $$,$$ as an electron travels from $$A$$ to $$B,$$ it slows from $${u_A} = 6.1 \times {10^6}\,m/s$$ to $${u_B} = 4.5 \times {10^6}\,m/s.$$ What is its potential change $$\Delta V = {V_B} - {V_A}$$ in volts?

The charge on each capacitors shown in figure and the potential difference across them will be respectively.