IIT JEE Main 2016 Physics Question 20

IIT JEE Main 2016 Physics Question No 20

A combination of capacitors is set up as shown in the figure.

IIT JEE Main 2016 Physics Question No 20

The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9μF capacity), at a point distant 30 m from it, would equal:

(1) 420 N/C (2) 480 N/C (3) 240 N/C (4) 360 N/C

Solution

Let V1 be the potential drop across 4μF capacitor and V2 across the parallel combination of 3μF and 9μF capacitors.
Now, from circuit diagram we have,
V1 + V2 = 8V and
4V1 = 12V2
On solving, V1 = 6 V, V2 = 2 V
Hence, charges Q1 = 24 μC, Q2 = 18 μc
Total charge Q = 42 μC
Hence, Electric Field E = (1/(4Πε0))(Q/r2)
=> E = (9 x 109 x 42 x 10-6)/302
=> E = 420 N/C

Answer

Correct answer to given problem No 20 of IIT JEE Main Physics paper of 2016 is option number (1) that is 420 N/C.

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IIT JEE Main 2016 Physics Question No 10

IIT JEE Main 2016 Physics Question No 10

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure ), have volume charge density ρ = A/r where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:

IIT JEE Main 2016 Physics Question No 10
(1) 2Q/(Π(a2-b2))
(2) 2Q/(Π(α2)
(3) Q/(2Π(α2)
(4) Q/(2Π(b2-a2))

Solution

Using Gauss Theorem for a Gaussian surface for radius r, we have

E 4Πr2 = (1/ε0)[Q + ∫ar(A/r)4Πr2dr]
E = (Q/(4Πε0r2)) + [(1/(4Πε0r2))2ΠA{r2 – ar2}]
E = (Q/(4Πε0r2)) + (A/2ε0) – (Aa2/2r2ε0)

for E to be independent of r

(Q/(4Πε0r2)) – (Aa2/2r2ε0) = 0

=> A = Q/2Πa2

Answer

Correct answer to given problem No 10 of IIT JEE Main Physics paper of 2016 is A = Q/2Πa2 i.e. option number 3.

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