Dear Student, We know that ϕ=∫E dSand dS=dxdyk^so ϕ=∬E dS=∬E0xdxdy=E0∫0ady∫0axdx=E0(y)0ax220a=E0a32 Regard

Solve this:

Solve this: ION (1) : FLUX CALCULATION AND GAUSS'S LAW A square Of side •a • is tying in xy plane such that two of its sides are lying on the axis. If an electric field is applied on the square. The flux passing through the square is Eoa3 (B) (A) Eoa• 2 Eoa (C) 3 (D) 2 Shows a charge Q placed at the centre of open face of a cylinder as C.

Dear Student,

W e k n o w t h a t ϕ = \int E . d S a n d d S = d x d y \hat{k} s o ϕ = \iint E . d S = \iint E_{0} x d x d y = E_{0} \int_{0}^{a} d y \int_{0}^{a} x d x = E_{0} (y)_{0}^{a} {[\frac{x^{2}}{2}]}_{0}^{a} = \frac{E_{0} a^{3}}{2}

Regard