## Sunday, October 02, 2011

### Multiple Choice Questions on Unusual Combinations of Capacitors

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err."

– Mahatma Gandhi

Today we will consider a few questions on unconventional combinations of capacitors which occasionally find place in entrance examination question papers conducted for admitting students to medical, engineering and other degree courses.

(1) Four parallel metal plates are arranged in air as shown, with the same separation ‘d’ between neighbouring plates. If the area (of one surface) of each plates is A, the capacitance between the terminals T1 and T2 is

(a) 4ε0A/d

(b) ε0A/d

(c) 2ε0A/d

(d) 3ε0A/d

(e) ε0A/3d

Let us number the plates starting from the top as 1, 2, 3, and 4 respectively. The lower surface of plate No.1 and the upper surface of plate No.2 with air in between makes a capacitor of capacitance ε0A/d.

The lower surface of plate No.2 and the upper surface of plate No.3 with air in between makes another capacitor of capacitance ε0A/d.

The lower surface of plate No.3 and the upper surface of plate No.4 with air in between makes a third capacitor of capacitance ε0A/d.

This arrangement therefore makes three capacitors in parallel. The capacitance value of each capacitor is ε0A/d so that the parallel combined value which appears between the terminals T1 and T2 is 3ε0A/d [Option (d)].

(2) If the connection of metal plates in the above question is modified as shown in the adjoining figure, the capacitance between terminals T1 and T2 is

(a) ε0A/2d

(b) 3ε0A/2d

(c)0A/d

(d) 0A/d

(e) 0A/3d

In this case the capacitor at the centre (having capacitance ε0A/d) appears directly across the terminals T1 and T2. The upper and lower capacitors (each of value ε0A/d) are connected in series and this series combination having effective capacitance ε0A/2d also appears between the terminals T1 and T2. Therefore, the effective capacitance between T1 and T2 is ε0A/d + ε0A/2d = 3ε0A/2d [Option (b)].

(3) If a fifth plate is added to the combination shown in question No.1, as shown in the figure, what is the effective capacitance between T1 and T2?

(a) ε0A/4d

(b) ε0A/5d

(c) 0A/d

(d) 0A/5d

(e) 0A/d

Once you answer question No.1, this question becomes quite simple. You can easily see that there are four identical capacitors (each having capacitance ε0A/d) connected in parallel. Therefore, the effective capacitance between T1 and T2 is 4ε0A/d [Option (e)].

You can find a few more questions (with solution) in this section here.