First slide

Gauss's law and its applications

Question

Three charges $q_{1} = 1 \times 10^{- 6} C, q_{2} = 2 \times 10^{- 6} C and q_{3} = - 3 \times 10^{- 6} C$ have been placed as shown in figure. Then the net electric flux will be maximum for the surface

Easy

A

S₁
B

S₂
C

S₃
D

same for all these

Solution

Charge inside $S_{1} = q_{1} + q_{2} = 3 \times 10^{- 6} C$
Charge inside $S_{2} = q_{2} + q_{3} = - 1 \times 10^{- 6} C$
Charge inside $S_{3} = 0$
Charge inside S₁ is greatest. So, flux through,S₁ is maximum.

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