Question

45. A point charge is placed at a distance
perpendicular to the plane and above the centre
of a square of side a. The electric flux through the
square is :-​

Answer 1

⭐《ANSWER》

$\huge\mathfrak{\underline {\underline\pink {ANSWER}}}$

↪Actually welcome to the concept of the ELECTRIC FLUX

↪According to the Gauss Law , we know that

↪The Electric Flux by a charge in a Closed surface is given by

↪FLUX = Q/Epsilon°

↪So basically here , the given condition is , the Square plate of side a and is horizontal

↪Thus the charge is placed perpendicular above the centre of the square plate ,

↪Thus it is 1/6 th part of a complete closed cube , so the flux will also be the 1/6 th strength of the closed cube

↪so the Electric flux through the plate will be

〽FLUX = Q/ 6 EPSILON°

Answer 2

Answer:

GAUSS 'S THEOREM,

By Gauss's theorem, total electric flux linked with a closed surface is given by

$\boxed{\mathsf{\Phi{\: = \:{\dfrac{q} {{\epsilon}_{0}}}}}}$

Where, q is the total charge enclosed by the closed surface.

As point charge is placed at a distance perpendicular to the plane and above centre of Square of side a, Electric flux will be symmetrically distributed through all faces.

Electric Flux through the square, = $\mathsf{\dfrac{\phi}{6}}$

Electric Flux = $\boxed{\mathsf{\dfrac{q}{6\:{\epsilon}_{0}}}}$