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JEE Main 2016 · 3 Apr · Shift 1, Physics question 13

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal :

An 8 V battery across two parallel branches: a 2 μF capacitor, and a 4 μF capacitor in series with 3 μF and 9 μF capacitors in parallel.
Choose one option
Answer and worked solution

Official answer: C

3 μF and 9 μF in parallel make 12 μF, and in series with 4 μF that gives 3 μF across 8 V: 24 μC. The 4 μF capacitor holds 24 μC. The 12 μF pair is at 2 V, so the 9 μF holds 18 μC. Then Q = 42 μC and
E=9×109×42×10−6302=420 N/C.E=\frac{9\times10^9\times42\times10^{-6}}{30^2}=420\text{ N/C}.

Topics:Series and parallel combinationsElectric field
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