Question

Take the potential of the point B in figure (31-E7) to be zero. (a) Find the potentials at the points C and D. (b) If a capacitor is connected between C and D, what charge will appear on this capacitor?
Figure

Answer 1

a) The potentials at the points C & D is 2 μF

b) The Charge appear on the capacitor is Zero.

Explanation:

Step 1:

Capacitance at point A is equal to  $C=(4 \times 8) /(4+8)$

Which is equal to 2 μF.

Where Q = Charge

C = Capacitance

V = Voltage in the circuit

$\therefore Q=8 / 3 \times 50=400 / 3$

Step: 2

Therefore the voltage drop at $4 \mu F=400 / 3 \times 4=100 / 3$

The voltage drop at $8 \mu F=400 / 3 \times 8=100 / 6$

The Potential difference is equal to the difference between the voltage drop across the two points

$100 / 3-100 / 6=50 \text { micro volts }$

Step 3:

The charge at Capacitor 2μF  

$50 \times 2=100 \mu \mathrm{F}$

Potential at $3 \mu F=100 / 3$

Potential at $6 \mu F=100 / 6$

Step 4:

Difference between the two potentials = $100 / 3-100 / 6=50 / 3 \mu V$

From the above points we can say that potential difference between the two points are same hence  the Charge appear on the capacitor is Zero.