A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is -

Question

A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-3cv83qcmvqyeeksg.jpg" /> (A) (Q/8 ∈ 0) (B) (Q/4 ∈ 0) (C) (Q/∈ 0) (D) Zero

Tardigrade Admin · Accepted Answer

The electric field due to Q at any point of square plate will be along the plane of square therefore angle between electric field and area vector of plate is 90° .∴ = 0

Q. A point charge $Q$ is placed at the corner of a square plate of side a then the flux through the square plate is -

$\frac{Q}{8 \in _{0}}$

$\frac{Q}{4 \in _{0}}$

$\frac{Q}{\in _{0}}$

Zero

The electric field due to $Q$ at any point of square plate will be along the plane of square therefore angle between electric field and area vector of plate is $9 0^{\circ} . ∴ = 0$

Q. A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is -

8∈0​Q​

4∈0​Q​

∈0​Q​