**(1) An unusual type of capacitor is made using four identical plates P _{1}, P_{2}, P_{3 }and P_{4, }each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P_{2} and P_{4}. Wires soldered to the plates P_{1} and P_{3} serve as terminals T_{1} and T_{2} of the system. What is the effective capacitance between the terminals T_{1} and T_{2}?**

**(a) (2****ε _{0}A)/3d**

**(b) (3****ε _{0}A)/2d**

**(c) (3****ε _{0}A)/d**

**(e) (**

**ε**

_{0}*A*)/2*d*The arrangement contains 3 identlcal capacitors C_{1}, C_{2} and C_{3} each of capacitance (ε_{0}*A*)/*d* arranged as shown in the adjoining figure. Plate P_{2} is common for C_{1} and C_{2}. Likewise, plates P_{3 }is common for C_{2} and C_{3}._{ } The capacitors C_{2} and C_{3} are *connected in parallel* (to produce an effective value 2ε_{0}*A*/*d*) and this parallel combination is *connected in series* with the capacitor C_{1} of value ε_{0}*A*/*d* . The effective capacitance *C* between the terminals T_{1} and T_{2} is therfore given by

C = [(2ε_{0}*A*/*d*) ×(ε_{0}*A*/*d*)] / [(2ε_{0}*A*/*d*) +(ε_{0}*A*/*d*)].

Therefore, C = (2ε_{0}*A*)/3*d*

**(2) Four 2**

**μF capacitors and a 3 μF capacitor are connected as shown in the figure. The effective capacitance between the points A and B is**

**(a)**

**3 μF**

**(b) 2**** μF **

**(c) ****1.5 μF **

**(d) ****1 μF **

**(e) ****0.5 μF**** **

This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance** **between the points A and B is 2 μF.

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