Given,
$\int_\text{s}\text{E.dS}=0$
It means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
Now, from Gauss' law,
$\int_\text{S}\text{E.dS}=\frac{\text{q}}{\in_0}$
Where q is charge enclosed by the surface.
Now,
$\int_\text{s}\text{E.dS}=0$
q = 0 i.e., net charge enclosed by the surface must be zero.
Hence, all other charges must necessarily be outside the surface.
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In an electroplating experiment, m gm of silver is deposited when 4 ampere of current flows for 2 minute. The amount (in gm) of silver deposited by 6 ampere of current for 40 second will be
|
(a) 4 m |
(b) m/2 |
(c) m/4 |
(d) 2 m |
An N-type and P-type silicon can be obtained by doping pure silicon with
|
(a) Arsenic and Phosphorous |
(b) Indium and Aluminium |
|
(c) Phosphorous and Indium |
(d) Aluminium and Boron |
AB is a potentiometer wire of length 100 cm and its resistance is 10 ohms. It is connected in series with a resistance R = 40 ohms and a battery of e.m.f. 2 V and negligible internal resistance. If a source of unknown e.m.f. E is balanced by 40 cm length of the potentiometer wire, the value of E is
|
(a) 0.8 V |
(b) 1.6 V |
(c) 0.08 V |
(d) 0.16 V |
A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The e.m.f. of the cell needed to make the bulb glow at full intensity is
|
(a) 4.5 V |
(b) 1.5 V |
(c) 2.67 V |
(d) 13.5 V |