JEE Main · 2016 · 03 Apr 2016 · Offline · Capacitance

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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 μ F

and

9 μ F

capacitors), at a point distant 30 m from it, would equal :

(A) $360 N / C$
(B) $420 N / C$ ✓ Correct
(C) $480 N / C$
(D) $240 N / C$

Answer: (B)

Solution

Tracing the network, the

9 μ

F capacitor carries

18 μ

C and the

4 μ

F capacitor carries

24 μ

C, so the equivalent point charge is

Q = 24 + 18 = 42 μ

C. At

r = 30

m its field is

E = \frac{k Q}{r ^{2}} = \frac{9 \times 1 0 ^{9} \times 42 \times 1 0 ^{- 6}}{3 0 ^{2}} = 420

N/C.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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